Human Physiology

human-physiology-ai999x

Cardiac output, blood pressure, vascular resistance and Poiseuille flow, lung volumes, ventilation and the alveolar gas equation, oxygen content and binding, acid-base, renal clearance and eGFR, Nernst and Goldman-Hodgkin-Katz potentials, cable properties, body composition and metabolic rate (computed values, not medical advice).

Life Sciences Question Answering Sentence Similarity Feature Extraction

Get it: desktop, command line, Python Compute in your browser Hugging Face ↗GitHub ↗

96capabilities

7topics

96formulas typeset

4ways to run: desktop app, command line, browser, Python

1.0.0version

What it does

What Human Physiology does

Human Physiology (human-physiology-ai999x) is an offline Python package for human physiology. It covers cardiac output, blood pressure, vascular resistance and Poiseuille flow, lung volumes, ventilation and the alveolar gas equation, oxygen content and binding, acid-base, renal clearance and eGFR, Nernst and Goldman-Hodgkin-Katz potentials, cable properties, body composition and metabolic rate (computed values, not medical advice).

It has 96 capabilities — 96 computational procedures — in 7 topics, the largest being Respiratory, Cardiovascular, Nerve and muscle, Renal, Metabolism and Body fluids. Ask a question in words with ask(), or call a capability by name with typed inputs with solve(). Every answer carries its value and unit, the steps that produced it and the formula in LaTeX.

When a question is not determined — a value is missing, an input is out of range, or the request is outside the package — it says so and names what it needs, instead of guessing. It runs on an ordinary CPU without a network connection (Python >=3.10), and the same question always gets the same answer.

What it covers

Body composition5Body density and fat by hydrostatic weighing · Body fat from body density (Siri and Brozek) · Body mass index …Body fluids6Body fluid compartments · Gibbs-Donnan equilibrium with an impermeant anion · Osmolarity of a solution …Cardiovascular26Bernoulli pressure drop across a narrowing · Cardiac index and stroke volume index · Cardiac output by the Fick principle …Metabolism8Basal metabolic rate by the Harris-Benedict equation · Basal metabolic rate by the Katch-McArdle equation · Energy expenditure of an activity from METs …Nerve and muscle12Charge and number of ions to change the membrane potential · Conduction velocity scaling with fibre diameter · Hill muscle force-velocity relation …Renal12Calculated and effective plasma osmolality and the osmolal gap · Creatinine clearance by the Cockcroft-Gault equation · Creatinine clearance from a timed urine collection …Respiratory27Alveolar gas equation · Alveolar ventilation from CO2 output and arterial PCO2 · Alveolar-arterial oxygen gradient, a/A ratio and P/F ratio …

Get it

Use Human Physiology your way

Use it right here in your browser now: the compiled program runs on your own device, with nothing sent anywhere. The AI999X desktop app for Windows and Linux and the command-line program human-physiology follow in their next releases.

Desktop app

One app with every AI999X model: search, ask in words, input forms, live figures and explanations. Human Physiology joins it in its next release; until then, compute with it in your browser.

The desktop app today · Compute in your browser

Command line

One program per model, for terminals and scripts. This model's program, human-physiology, joins the command-line bundle in its next release.

What it answers, command by command

Python

The Python package, for notebooks and your own programs (Python 3.10 or newer).

pip install human-physiology-ai999x

Examples · Python guide

From here to first answer

The browser works now; the downloads follow with the next release.

Type a question in words in the calculator below, or choose a capability (or open any capability further down).
Its inputs come filled in with a worked example: change any of them, in any unit of the right kind.
Press Compute: the answer, what it means and its figures, computed on your device.

Open the calculator

In your browser

Compute something now

Ask a question in words, or choose a capability and open its calculator: the inputs come filled in with a worked example; change any of them (any unit of the right kind) and press Compute. It runs on your device; nothing is sent anywhere.

Capability

The same capabilities as the desktop app and the command line, as WebAssembly in this page. Every capability below also has its own Compute in your browser button.

Worked examples

One example, every way to run it

Each worked example with its answer, then the same question on the command line, from Python and in your browser. A command-line output is exactly what the program printed for that command.

Cardiovascular

cardiac output: HR = 72, SV = 70 mL

CO = HR \times SV

CO = 5.04

CO_mL = 5040

How it was worked out

Cardiac output from heart rate and stroke volume
CO = HR \times SV
with HR = 72, SV = 70 mL
→ CO = 5.04, CO_mL = 5040

Terminal

$ human-physiology solve hp_cardiac_output HR=72 "SV=70 mL"
CO = 5.04
CO_mL = 5040
  by: Cardiac output from heart rate and stroke volume  (CO = HR SV / 1000)
  HR = 72
  SV = 7e-05 m^3
  assumes: HR > 0; SV > 0

import human_physiology_ai999x as m

a = m.ask("cardiac output: HR = 72, SV = 70 mL")
print(a.values or (a.value, a.unit))

About this capability

Respiratory

alveolar gas equation: FiO2 = 21%, Patm = 760 mmHg, PH2O = 47 mmHg, PaCO2 = 40 mmHg, RQ = 0.8

P_AO_2 = F_iO_2(P_B – P_{H_2O}) – \frac{P_aCO_2}{R}\left[1 – F_iO_2(1 – R)\right]

PiO2 = 149.73

PAO2 = 99.73

PAO2_full = 101.83

How it was worked out

Alveolar gas equation
P_AO_2 = F_iO_2(P_B – P_{H_2O}) – \frac{P_aCO_2}{R}\left[1 – F_iO_2(1 – R)\right]
with FiO2 = 0.21, Patm = 760 mmHg, PH2O = 47 mmHg, PaCO2 = 40 mmHg, RQ = 0.8
→ PiO2 = 149.73, PAO2 = 99.73, PAO2_full = 101.83

Terminal

$ human-physiology solve hp_alveolar_gas_equation FiO2=0.21 "Patm=760 mmHg" "PH2O=47 mmHg" "PaCO2=40 mmHg" RQ=0.8
PiO2 = 149.73
PAO2 = 99.73
PAO2_full = 101.83
  by: Alveolar gas equation  (PAO2 = PiO2 - PaCO2/R (+ PaCO2 FiO2 (1 - R)/R))
  FiO2 = 0.21
  Patm = 101325.0144 Pa
  PH2O = 6266.152209 Pa
  PaCO2 = 5332.895497 Pa
  RQ = 0.8
  assumes: 0 < FiO2 <= 1; Patm > PH2O; 0 < RQ <= 1.5

import human_physiology_ai999x as m

a = m.ask("alveolar gas equation: FiO2 = 21%, Patm = 760 mmHg, PH2O = 47 mmHg, PaCO2 = 40 mmHg, RQ = 0.8")
print(a.values or (a.value, a.unit))

About this capability

Nerve and muscle

Nernst potential: z = 1, Cout = 4, Cin = 140

E = \frac{RT}{zF}\ln\frac{[C]_o}{[C]_i}

E = -95.02257566

RT_F = 26.72665911

How it was worked out

Nernst equilibrium potential of an ion
E = \frac{RT}{zF}\ln\frac{[C]_o}{[C]_i}
with z = 1, Cout = 4, Cin = 140
→ E = -95.02257566, RT_F = 26.72665911

Terminal

$ human-physiology solve hp_nernst_potential z=1 Cout=4 Cin=140
E = -95.02257566
RT_F = 26.72665911
  by: Nernst equilibrium potential of an ion  (Nernst equation)
  Cout = 4
  Cin = 140
  assumes: z a nonzero whole number; concentrations > 0

import human_physiology_ai999x as m

a = m.ask("Nernst potential: z = 1, Cout = 4, Cin = 140")
print(a.values or (a.value, a.unit))

About this capability

Cardiovascular

Poiseuille flow, mean velocity and wall shear stress in a vessel

Q = \frac{\pi r^4 \Delta P}{8 \eta L},\quad \tau_w = \frac{4\eta Q}{\pi r^3}

Q = 23.93397902

Q_mL_min = 1436.038741

v_mean = 190.4605534

tau_wall = 13.33223874

How it was worked out

Poiseuille flow, mean velocity and wall shear stress in a vessel
Q = \frac{\pi r^4 \Delta P}{8 \eta L},\quad \tau_w = \frac{4\eta Q}{\pi r^3}
with dP = 10 mmHg, eta = 3.5, L = 10 cm, r = 2 mm
→ Q = 23.93397902, Q_mL_min = 1436.038741, v_mean = 190.4605534, tau_wall = 13.33223874

Terminal

$ human-physiology solve hp_poiseuille_flow "dP=10 mmHg" eta=3.5 "L=10 cm" "r=2 mm"
Q = 23.93397902
Q_mL_min = 1436.038741
v_mean = 190.4605534
tau_wall = 13.33223874
  by: Poiseuille flow, mean velocity and wall shear stress in a vessel  (Q = dP pi r^4 /(8 eta L); tau_w = 4 eta Q/(pi r^3))
  dP = 1333.223874 Pa
  eta = 3.5
  L = 0.1 m
  r = 0.002 m
  assumes: all inputs > 0; steady laminar Newtonian flow

import human_physiology_ai999x as m

a = m.solve("hp_poiseuille_flow", dP="10 mmHg", eta=3.5, L="10 cm", r="2 mm")
print(a.values or (a.value, a.unit))

About this capability

Renal

eGFR CKD-EPI 2021: Scr = 1.0, age = 50, sex = male

eGFR = 142\,\min(S_{cr}/\kappa,1)^{\alpha}\max(S_{cr}/\kappa,1)^{-1.200}\,0.9938^{age}\,[1.012]

eGFR = 91.69147861

How it was worked out

Estimated GFR by the CKD-EPI 2021 creatinine equation
eGFR = 142\,\min(S_{cr}/\kappa,1)^{\alpha}\max(S_{cr}/\kappa,1)^{-1.200}\,0.9938^{age}\,[1.012]
with Scr = 1, age = 50, sex = male
→ eGFR = 91.69147861

Terminal

$ human-physiology solve hp_egfr_ckd_epi_2021 Scr=1 age=50 sex=male
eGFR = 91.69147861
  by: Estimated GFR by the CKD-EPI 2021 creatinine equation  (CKD-EPI 2021: k = 0.7/0.9, a = -0.241/-0.302 (female/male))
  Scr = 1
  age = 50
  sex = male
  assumes: Scr > 0; 18 <= age <= 120

import human_physiology_ai999x as m

a = m.ask("eGFR CKD-EPI 2021: Scr = 1.0, age = 50, sex = male")
print(a.values or (a.value, a.unit))

About this capability

Body composition

body mass index: weight = 70 kg, height = 175 cm

BMI = \frac{W}{H^2}

BMI = 22.85714286

How it was worked out

Body mass index
BMI = \frac{W}{H^2}
with weight = 70 kg, height = 175 cm
→ BMI = 22.85714286

Terminal

$ human-physiology solve hp_bmi "weight=70 kg" "height=175 cm"
BMI = 22.85714286
  by: Body mass index  (BMI = W / H^2 (H in m))
  weight = 70 kg
  height = 1.75 m
  assumes: weight, height > 0

import human_physiology_ai999x as m

a = m.ask("body mass index: weight = 70 kg, height = 175 cm")
print(a.values or (a.value, a.unit))

About this capability

Body fluids

body fluid compartments: weight = 70 kg

TBW = 0.6\,W,\; ICF = \tfrac23 TBW,\; ECF = \tfrac13 TBW

TBW = 42

ICF = 28

ECF = 14

ISF = 10.5

plasma = 3.5

How it was worked out

Body fluid compartments
TBW = 0.6\,W,\; ICF = \tfrac23 TBW,\; ECF = \tfrac13 TBW
with weight = 70 kg
→ TBW = 42, ICF = 28, ECF = 14, ISF = 10.5, plasma = 3.5

Terminal

$ human-physiology solve hp_body_fluid_compartments "weight=70 kg"
TBW = 42
ICF = 28
ECF = 14
ISF = 10.5
plasma = 3.5
  by: Body fluid compartments  (60-40-20 rule: ICF 2/3, ECF 1/3 of TBW; plasma 1/4 of ECF)
  weight = 70 kg
  assumes: weight > 0

import human_physiology_ai999x as m

a = m.ask("body fluid compartments: weight = 70 kg")
print(a.values or (a.value, a.unit))

About this capability

Metabolism

Atwater energy: carbohydrate = 300 g, fat = 70 g, protein = 80 g

E = 4C + 9F + 4P + 7A

kcal = 2150

kJ = 8995.6

How it was worked out

Food energy by the Atwater factors
E = 4C + 9F + 4P + 7A
with carbohydrate = 300 g, fat = 70 g, protein = 80 g
→ kcal = 2150, kJ = 8995.6

Terminal

$ human-physiology solve hp_atwater_energy "carbohydrate=300 g" "fat=70 g" "protein=80 g"
kcal = 2150
kJ = 8995.6
  by: Food energy by the Atwater factors  (4/9/4/7 kcal per g; 1 kcal = 4.184 kJ)
  carbohydrate = 0.3 kg
  fat = 0.07 kg
  protein = 0.08 kg
  assumes: amounts >= 0, not all zero

import human_physiology_ai999x as m

a = m.ask("Atwater energy: carbohydrate = 300 g, fat = 70 g, protein = 80 g")
print(a.values or (a.value, a.unit))

About this capability

Command line

Use it from the command line

Every command below was run as shown, and the text under it is exactly what human-physiology printed. The same commands work in a Linux terminal and in Windows PowerShell. human-physiology joins the command-line bundle in its next release. The user guide covers install, inputs, outcomes, scripting, and every capability with a command and what it printed: PDF · text.

Check the install

Prints the model, its version and that this is the Rust build.

Terminal

$ human-physiology --version
human-physiology-ai999x 1.0.0 (Rust build, ax-engine 0.1.0)

Ask in words

Write the question as you would say it: the program reads the quantities and their units, picks the capability and answers.

Terminal

$ human-physiology ask "cardiac output: HR = 72, SV = 70 mL"
Read as: Cardiac output from heart rate and stroke volume
CO = 5.04
CO_mL = 5040
  by: Cardiac output from heart rate and stroke volume  (CO = HR SV / 1000)
  HR = 72
  SV = 7e-05 m^3
  assumes: HR > 0; SV > 0

Find a capability

Search the capabilities by what you want to know; the first column is the id to pass to info and solve.

Terminal

$ human-physiology search "cardiac output from heart rate and stroke volume"
hp_cardiac_output                    Cardiac output from heart rate and stroke volume  (cardiovascular / cardiac output)
hp_cardiac_index                     Cardiac index and stroke volume index  (cardiovascular / cardiac output)
hp_stroke_work                       Stroke work and cardiac power  (cardiovascular / cardiac work)
hp_stroke_volume_ef                  Stroke volume and ejection fraction  (cardiovascular / cardiac output)
hp_qt_correction                     Heart-rate corrected QT interval (Bazett, Fridericia, Framingham)  (cardiovascular / electrocardiogram)
hp_heart_rate_reserve                Maximal heart rate, heart rate reserve and Karvonen target  (metabolism / exercise)
hp_fick_cardiac_output               Cardiac output by the Fick principle  (cardiovascular / cardiac output)
hp_fick_saturations                  Fick cardiac output from hemoglobin, saturations and PO2  (cardiovascular / cardiac output)
… 12 more

List the capabilities

Every capability the program answers: its id (what you pass to solve) and its name.

Terminal

$ human-physiology list
hp_aa_gradient                           Alveolar-arterial oxygen gradient, a/A ratio and P/F ratio
hp_alveolar_gas_equation                 Alveolar gas equation
hp_alveolar_laplace_pressure             Laplace pressure of an alveolus
hp_alveolar_ventilation_co2              Alveolar ventilation from CO2 output and arterial PCO2
hp_anion_gap                             Serum anion gap and albumin-corrected anion gap
hp_atwater_energy                        Food energy by the Atwater factors
hp_barometric_altitude                   Barometric pressure and inspired PO2 at altitude
hp_bernoulli_pressure_drop               Bernoulli pressure drop across a narrowing
… 88 more

What a capability needs

The equation or method, every input and output with its SI unit. Read this before solve: it tells you the input names.

Terminal

$ human-physiology info hp_cardiac_output
Cardiac output from heart rate and stroke volume
  CO = HR SV / 1000
  in   HR       heart rate in beats/min []
  in   SV       stroke volume (e.g. in mL) [m^3]
  out  CO       cardiac output in L/min []
  out  CO_mL    cardiac output in mL/min []

Solve

Inputs as name=value, each with any unit of the right kind. The answer lists every output, the method used, each input converted to SI, and the conditions the answer assumes.

Terminal

$ human-physiology solve hp_cardiac_output HR=72.0 "SV=70 mL"
CO = 5.04
CO_mL = 5040
  by: Cardiac output from heart rate and stroke volume  (CO = HR SV / 1000)
  HR = 72
  SV = 7e-05 m^3
  assumes: HR > 0; SV > 0

Machine-readable answer

–json prints the same answer as one line of JSON (values in SI): for scripts and other programs.

Terminal

$ human-physiology solve hp_cardiac_output HR=72.0 "SV=70 mL" --json
{"outcome":"SOLVED","message":"","value":5.04,"unit":"","solved_for":"CO","si":[["HR",72.0],["SV",0.00007]],"values":[["CO",5.04,""],["CO_mL",5040.0,""]],"target":"CO","phrase":"cardiac output in L/min","assumptions":["HR > 0","SV > 0"],"name":"Cardiac output from heart rate and stroke volume","equation":"CO = HR SV / 1000"}

When an input is missing

No number is guessed: the answer names what is missing, shows a complete call, and the program exits with status 1.

Terminal

$ human-physiology solve hp_cardiac_output HR=72.0
NEEDS_INPUT: hp_cardiac_output needs SV (stroke volume (e.g. in mL)). Example: solve('hp_cardiac_output', HR=72, SV='7e-05 m^3').

Many questions at once

One JSON request per line in, one JSON answer per line out, in order: for scripts, notebooks and agents (the program starts once). In PowerShell: Get-Content cases.jsonl | human-physiology batch

Terminal

cases.jsonl
{"key": "hp_cardiac_output", "inputs": {"HR": 72.0, "SV": "70 mL"}}
{"key": "hp_alveolar_gas_equation", "inputs": {"FiO2": 0.21, "Patm": "760 mmHg", "PH2O": "47 mmHg", "PaCO2": "40 mmHg", "RQ": 0.8}}

$ human-physiology batch < cases.jsonl
{"outcome":"SOLVED","message":"","value":5.04,"unit":"","solved_for":"CO","si":[["HR",72.0],["SV",0.00007]],"values":[["CO",5.04,""],["CO_mL",5040.0,""]],"target":"CO","phrase":"cardiac output in L/min","assumptions":["HR > 0","SV > 0"],"name":"Cardiac output from heart rate and stroke volume","equation":"CO = HR SV / 1000"}
{"outcome":"SOLVED","message":"","value":99.73,"unit":"","solved_for":"PAO2","si":[["FiO2",0.21],["Patm",101325.01443540001],["PH2O",6266.152208505],["PaCO2",5332.8954966],["RQ",0.8]],"values":[["PiO2",149.73,""],["PAO2",99.73,""],["PAO2_full",101.83,""]],"target":"PAO2","phrase":"inspired PO2 in mmHg","assumptions":["0 < FiO2 <= 1","Patm > PH2O","0 < RQ <= 1.5"],"name":"Alveolar gas equation","equation":"PAO2 = PiO2 - PaCO2/R (+ PaCO2 FiO2 (1 - R)/R)"}

All commands

Command	What it does
`human-physiology ask "<question>"`	answer a question written in words
`human-physiology search "<words>"`	find the capabilities that match
`human-physiology list`	every capability: its id and name
`human-physiology info <capability>`	the equation or method, its inputs and outputs with their units
`human-physiology solve <capability> name=value …`	the answer; add –target X to solve an equation for X, –json for JSON
`human-physiology figures <capability> name=value … --out DIR`	the live figures for those inputs, as SVG files in DIR
`human-physiology batch < requests.jsonl`	one JSON answer per line, for scripts and agents
`human-physiology catalog`	every capability with its inputs, as JSON
`human-physiology --version`	the model and its version

Exit status: 0 when the question was answered, 1 when it was not (the message says why: a missing input, an invalid value, a question outside the model), 2 for a command it does not know.

Python

Use it from Python

After pip install human-physiology-ai999x, ask in words or call a capability. Every snippet was run as shown; the output under it is what it printed. The Python guide covers ask in words, solve, and every capability with a Python example: PDF · text.

import human_physiology_ai999x as m

a = m.ask("cardiac output: HR = 72, SV = 70 mL")
print(a.outcome)
print(a.values)
for step in a.steps:
    print(step["name"])

Output

SOLVED
{'CO': 5.04, 'CO_mL': 5040.0}
Cardiac output from heart rate and stroke volume

import human_physiology_ai999x as m

a = m.solve("hp_poiseuille_flow", dP="10 mmHg", eta=3.5, L="10 cm", r="2 mm")
print(a.outcome)
print(a.values)

Output

SOLVED
{'Q': 23.9339790207, 'Q_mL_min': 1436.03874124, 'v_mean': 190.46055345, 'tau_wall': 13.3322387415}

import human_physiology_ai999x as m

# an input is missing: the answer says what it needs instead of guessing
a = m.solve("hp_poiseuille_flow")
print(a.outcome)
print(a.message)

Output

NEEDS_INPUT
hp_poiseuille_flow needs dP, eta, L, r (pressure drop along the vessel (e.g. in mmHg); blood viscosity in cP (mPa s); vessel length (e.g. in cm); vessel radius (e.g. in mm)). Example: solve('hp_poiseuille_flow', dP='1333.22 Pa', eta=3.5, L='0.1 m', r='0.002 m').

human-physiology-ai999x "cardiac output: HR = 72, SV = 70 mL"

Output

CO = 5.04, CO_mL = 5040
1. Cardiac output from heart rate and stroke volume: CO = HR SV / 1000  with HR = 72, SV = 70 mL  ->  CO = 5.04, CO_mL = 5040
Assumptions: HR > 0; SV > 0
…

Capabilities

Every capability, explained

Search by name, formula, input or topic. Each capability opens to say what it does, what it needs, what it gives back and why, with its worked example, how to run it (on the command line, from Python, in your browser) and its figures.

96 capabilities

Body composition 5

Body density and fat by hydrostatic weighing1 fig.hp_hydrostatic_weighing

Body volume (Ma – Mw)/rho_w – RV – GI gas, body density and Siri percent fat.

D_b = \frac{M_a}{\frac{M_a – M_w}{\rho_w} – RV – V_{GI}}

You give

	Meaning	Unit	Example
\text{mass\_air}	Body mass in air (e.g. in kg)	kg	70 kg
\text{mass\_water}	Underwater mass (e.g. in kg)	kg	3 kg
\text{water\_density}	Water density (e.g. in g/mL)	kg/m^3	0.9957 g/mL
\text{RV}	Residual lung volume (e.g. in L)	m^3	1.2 L
\text{GI}	Gastrointestinal gas (e.g. in L) (optional; the customary estimate if omitted) optional	m^3

You get

	Meaning	Unit	Example
\text{volume}	Body volume in L	—	65.98934418
\text{density}	Body density in g/mL	—	1.060777325
\text{fat\_Siri}	Body fat in % (Siri)	—	16.63893384

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

mass_water < mass_air
computed density 0.9-1.2

Outside these conditions it says what is wrong instead of returning a number.

Worked example

hydrostatic weighing: mass_air = 70 kg, mass_water = 3 kg, water_density = 0.9957 g/mL, RV = 1.2 L

volume = 65.98934418

density = 1.060777325

fat_Siri = 16.63893384

Run it

Terminal

$ human-physiology solve hp_hydrostatic_weighing "mass_air=70 kg" "mass_water=3 kg" "water_density=0.9957 g/mL" "RV=1.2 L"
volume = 65.98934418
density = 1.060777325
fat_Siri = 16.63893384
  by: Body density and fat by hydrostatic weighing  (Archimedes' principle, then Siri)
  mass_air = 70 kg
  mass_water = 3 kg
  water_density = 995.7 kg/m^3
  RV = 0.0012 m^3
  assumes: mass_water < mass_air; computed density 0.9-1.2

import human_physiology_ai999x as m

a = m.solve("hp_hydrostatic_weighing", mass_air="70 kg", mass_water="3 kg", water_density="0.9957 g/mL", RV="1.2 L")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Body fat from body density (Siri and Brozek)2 fig.hp_body_fat_from_density

Percent body fat by Siri (495/D – 450) and Brozek (457/D – 414.2) from whole-body density.

\%BF = \frac{495}{D} – 450

You give

	Meaning	Unit	Example
\text{density}	Body density (e.g. in g/mL)	kg/m^3	1.05 g/mL

You get

	Meaning	Unit	Example
\text{fat\_Siri}	Body fat in % (Siri)	—	21.42857143
\text{fat\_Brozek}	Body fat in % (Brozek)	—	21.03809524

Values with a unit may be written in any unit of the right kind.

Applies when

0.9 < density < 1.2

Outside these conditions it says what is wrong instead of returning a number.

Worked example

body fat from density: density = 1.05 g/mL

fat_Siri = 21.42857143

fat_Brozek = 21.03809524

Run it

Terminal

$ human-physiology solve hp_body_fat_from_density "density=1.05 g/mL"
fat_Siri = 21.42857143
fat_Brozek = 21.03809524
  by: Body fat from body density (Siri and Brozek)  (two-compartment models)
  density = 1050 kg/m^3
  assumes: 0.9 < density < 1.2

import human_physiology_ai999x as m

a = m.solve("hp_body_fat_from_density", density="1.05 g/mL")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Body mass index2 fig.hp_bmi

BMI = weight/height^2 (kg/m^2).

BMI = \frac{W}{H^2}

You give

	Meaning	Unit	Example
\text{weight}	Weight (e.g. in kg)	kg	70 kg
\text{height}	Height (e.g. in cm)	m	175 cm

You get

	Meaning	Unit	Example
\text{BMI}	Body mass index in kg/m²	—	22.85714286

Values with a unit may be written in any unit of the right kind.

Applies when

weight, height > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

body mass index: weight = 70 kg, height = 175 cm

BMI = 22.85714286

Run it

Terminal

$ human-physiology solve hp_bmi "weight=70 kg" "height=175 cm"
BMI = 22.85714286
  by: Body mass index  (BMI = W / H^2 (H in m))
  weight = 70 kg
  height = 1.75 m
  assumes: weight, height > 0

import human_physiology_ai999x as m

a = m.solve("hp_bmi", weight="70 kg", height="175 cm")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Body surface area (DuBois, Mosteller, Haycock, Gehan-George)2 fig.hp_body_surface_area

Body surface area by the DuBois (1916), Mosteller (1987), Haycock (1978) and Gehan-George (1970) formulas.

BSA_{Mosteller} = \sqrt{\frac{H\,W}{3600}},\quad BSA_{DuBois} = 0.007184\,W^{0.425}H^{0.725}

You give

	Meaning	Unit	Example
\text{weight}	Weight (e.g. in kg)	kg	70 kg
\text{height}	Height (e.g. in cm)	m	175 cm

You get

	Meaning	Unit	Example
\text{BSA\_DuBois}	DuBois BSA in m²	—	1.848143018
\text{BSA\_Mosteller}	Mosteller BSA in m²	—	1.844661968
\text{BSA\_Haycock}	Haycock BSA in m²	—	1.846776349
\text{BSA\_GehanGeorge}	Gehan-George BSA in m²	—	1.853853298

Values with a unit may be written in any unit of the right kind.

Applies when

weight, height > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

body surface area: weight = 70 kg, height = 175 cm

BSA_DuBois = 1.848143018

BSA_Mosteller = 1.844661968

BSA_Haycock = 1.846776349

BSA_GehanGeorge = 1.853853298

Run it

Terminal

$ human-physiology solve hp_body_surface_area "weight=70 kg" "height=175 cm"
BSA_DuBois = 1.848143018
BSA_Mosteller = 1.844661968
BSA_Haycock = 1.846776349
BSA_GehanGeorge = 1.853853298
  by: Body surface area (DuBois, Mosteller, Haycock, Gehan-George)  (published power-law formulas)
  weight = 70 kg
  height = 1.75 m
  assumes: weight, height > 0

import human_physiology_ai999x as m

a = m.solve("hp_body_surface_area", weight="70 kg", height="175 cm")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Lean body mass by the Boer formula2 fig.hp_lean_body_mass_boer

Boer (1984): male 0.407 W + 0.267 H – 19.2, female 0.252 W + 0.473 H – 48.3 (kg), and the remaining fat mass.

LBM_m = 0.407\,W + 0.267\,H – 19.2

You give

	Meaning	Unit	Example
\text{weight}	Weight (e.g. in kg)	kg	70 kg
\text{height}	Height (e.g. in cm)	m	175 cm
\text{sex}	Sex for the formula choice	—	male

You get

	Meaning	Unit	Example
\text{LBM}	Lean body mass in kg	—	56.015
\text{fat\_mass}	Weight – LBM in kg	—	13.985

Values with a unit may be written in any unit of the right kind.

Applies when

0 < LBM < weight

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Boer lean body mass: weight = 70 kg, height = 175 cm, sex = male

LBM = 56.015

fat_mass = 13.985

Run it

Terminal

$ human-physiology solve hp_lean_body_mass_boer "weight=70 kg" "height=175 cm" sex=male
LBM = 56.015
fat_mass = 13.985
  by: Lean body mass by the Boer formula  (Boer anthropometric formula)
  weight = 70 kg
  height = 1.75 m
  sex = male
  assumes: 0 < LBM < weight

import human_physiology_ai999x as m

a = m.solve("hp_lean_body_mass_boer", weight="70 kg", height="175 cm", sex="male")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Body fluids 6

Body fluid compartments1 fig.hp_body_fluid_compartments

Total body water (60 % of weight by default), intracellular 2/3, extracellular 1/3, interstitial 3/4 and plasma 1/4 of the ECF.

TBW = 0.6\,W,\; ICF = \tfrac23 TBW,\; ECF = \tfrac13 TBW

You give

	Meaning	Unit	Example
\text{weight}	Body weight (e.g. in kg)	kg	70 kg
\text{TBW\_fraction}	Total body water fraction of weight (optional; the textbook adult fraction if omitted) optional	—

You get

	Meaning	Unit	Example
\text{TBW}	Total body water in L	—	42
\text{ICF}	Intracellular fluid in L	—	28
\text{ECF}	Extracellular fluid in L	—	14
\text{ISF}	Interstitial fluid in L	—	10.5
\text{plasma}	Plasma volume in L	—	3.5

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

weight > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

body fluid compartments: weight = 70 kg

TBW = 42

ICF = 28

ECF = 14

ISF = 10.5

plasma = 3.5

Run it

Terminal

$ human-physiology solve hp_body_fluid_compartments "weight=70 kg"
TBW = 42
ICF = 28
ECF = 14
ISF = 10.5
plasma = 3.5
  by: Body fluid compartments  (60-40-20 rule: ICF 2/3, ECF 1/3 of TBW; plasma 1/4 of ECF)
  weight = 70 kg
  assumes: weight > 0

import human_physiology_ai999x as m

a = m.solve("hp_body_fluid_compartments", weight="70 kg")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Gibbs-Donnan equilibrium with an impermeant anion2 fig.hp_gibbs_donnan

Equilibrium NaCl distribution across a membrane when one side holds impermeant anions (charge concentration A) facing a large bath of NaCl c: concentrations, Donnan ratio, potential and osmotic excess.

[Na^+]_i[Cl^-]_i = [Na^+]_o[Cl^-]_o,\quad [Na^+]_i = [Cl^-]_i + A

You give

	Meaning	Unit	Example
A	Impermeant anion charge concentration inside in mEq/L	—	20
c	Bath NaCl concentration (e.g. in mmol/L)	mol/m^3	150 mmol/L
T	Temperature (e.g. in degC) (optional; body temperature if omitted) optional	K

You get

	Meaning	Unit	Example
\text{Na\_in}	Na+ inside in mmol/L	—	160.3329638
\text{Cl\_in}	Cl- inside in mmol/L	—	140.3329638
r	Donnan ratio Na_in/Na_out = Cl_out/Cl_in	—	1.068886425
V_{\mathrm{mV}}	Membrane potential inside relative to the bath in mV	—	-1.780460072
\text{osmotic\_excess}	Extra diffusible ions inside in mmol/L	—	0.6659275675

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

A, c > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Gibbs-Donnan equilibrium: A = 20, c = 150 mmol/L

Na_in = 160.3329638

Cl_in = 140.3329638

r = 1.068886425

V_mV = -1.780460072

osmotic_excess = 0.6659275675

Run it

Terminal

$ human-physiology solve hp_gibbs_donnan A=20 "c=150 mmol/L"
Na_in = 160.3329638
Cl_in = 140.3329638
r = 1.068886425
V_mV = -1.780460072
osmotic_excess = 0.6659275675
  by: Gibbs-Donnan equilibrium with an impermeant anion  (Na_in Cl_in = c^2 with Na_in = Cl_in + A (quadratic); V = -(RT/F) ln(Na_in/c))
  A = 20
  c = 150 mol/m^3
  assumes: A, c > 0

import human_physiology_ai999x as m

a = m.solve("hp_gibbs_donnan", A="20", c="150 mmol/L")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Osmolarity of a solution1 fig.hp_solution_osmolarity

Molarity from a mass concentration and the osmolarity (particles per formula unit, times the osmotic coefficient).

\text{osmolarity} = \phi\, i\, \frac{c}{M}

You give

	Meaning	Unit	Example
\text{conc}	Mass concentration (e.g. in g/L)	kg/m^3	9 g/L
\text{molar\_mass}	Molar mass (e.g. in g/mol)	kg/mol	58.44 g/mol
i	Number of particles per formula unit (dissociation)	—	2
\phi	Osmotic coefficient (optional; ideal solution if omitted) optional	—

You get

	Meaning	Unit	Example
\text{molarity}	Molarity in mmol/L	—	154.0041068
\text{osmolarity}	Osmolarity in mOsm/L	—	308.0082136

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

conc, molar mass > 0
1 <= i <= 6

Outside these conditions it says what is wrong instead of returning a number.

Worked example

solution osmolarity: conc = 9 g/L, molar_mass = 58.44 g/mol, i = 2

molarity = 154.0041068

osmolarity = 308.0082136

Run it

Terminal

$ human-physiology solve hp_solution_osmolarity "conc=9 g/L" "molar_mass=58.44 g/mol" i=2
molarity = 154.0041068
osmolarity = 308.0082136
  by: Osmolarity of a solution  (mOsm/L = 1000 c/M x i x phi)
  conc = 9 kg/m^3
  molar_mass = 0.05844 kg/mol
  assumes: conc, molar mass > 0; 1 <= i <= 6

import human_physiology_ai999x as m

a = m.solve("hp_solution_osmolarity", conc="9 g/L", molar_mass="58.44 g/mol", i="2")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Osmotic pressure by van 't Hoff's law1 fig.hp_osmotic_pressure

pi = c R T for an osmolarity c, in Pa, mmHg and atm (about 19.3 mmHg per mOsm/L at 37 degC).

\pi = c R T

You give

	Meaning	Unit	Example
\text{osm}	Osmolarity (or osmolarity difference) in mOsm/L	—	300
T	Temperature (e.g. in degC) (optional; body temperature if omitted) optional	K

You get

	Meaning	Unit	Example
\pi	Osmotic pressure in Pa	—	773619.1743
\pi_{\mathrm{mmHg}}	Osmotic pressure in mmHg	—	5802.62017
\pi_{\mathrm{atm}}	Osmotic pressure in atm	—	7.635027627

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

osm >= 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

osmotic pressure: osm = 300

pi = 773619.1743

pi_mmHg = 5802.62017

pi_atm = 7.635027627

Run it

Terminal

$ human-physiology solve hp_osmotic_pressure osm=300
pi = 773619.1743
pi_mmHg = 5802.62017
pi_atm = 7.635027627
  by: Osmotic pressure by van 't Hoff's law  (pi = c R T (ideal dilute solution))
  osm = 300
  assumes: osm >= 0

import human_physiology_ai999x as m

a = m.solve("hp_osmotic_pressure", osm="300")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Total body water by the Watson equations1 fig.hp_watson_total_body_water

Watson (1980): male 2.447 – 0.09516 age + 0.1074 H + 0.3362 W; female -2.097 + 0.1069 H + 0.2466 W (litres).

TBW_{m} = 2.447 – 0.09516\,A + 0.1074\,H + 0.3362\,W

You give

	Meaning	Unit	Example
\text{age}	Age in years	—	40
\text{height}	Height (e.g. in cm)	m	175 cm
\text{weight}	Weight (e.g. in kg)	kg	70 kg
\text{sex}	Sex for the equation choice	—	male

You get

	Meaning	Unit	Example
\text{TBW}	Total body water in L	—	40.9696

Values with a unit may be written in any unit of the right kind.

Applies when

adult
height, weight > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Watson total body water: age = 40, height = 175 cm, weight = 70 kg, sex = male

TBW = 40.9696

Run it

Terminal

$ human-physiology solve hp_watson_total_body_water age=40 "height=175 cm" "weight=70 kg" sex=male
TBW = 40.9696
  by: Total body water by the Watson equations  (Watson anthropometric regression)
  age = 40
  height = 1.75 m
  weight = 70 kg
  sex = male
  assumes: adult; height, weight > 0

import human_physiology_ai999x as m

a = m.solve("hp_watson_total_body_water", age="40", height="175 cm", weight="70 kg", sex="male")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Two-compartment fluid shift after adding a solution (Darrow-Yannet)2 fig.hp_fluid_shift

New osmolality and intracellular and extracellular volumes at osmotic equilibrium after a volume of solution enters the ECF.

P_{osm}' = \frac{TBW\,P_{osm} + V C}{TBW + V},\quad ICF' = \frac{ICF\,P_{osm}}{P_{osm}'}

You give

	Meaning	Unit	Example
\text{TBW}	Initial total body water (e.g. in L)	m^3	42 L
\text{Posm}	Initial osmolality in mOsm/kg	—	290
V_{\mathrm{added}}	Volume added to the ECF (e.g. in L)	m^3	2 L
C_{\mathrm{added}}	Osmolarity of the added solution in mOsm/L	—	0
\text{ICF\_fraction}	Initial intracellular fraction of total body water (optional; the textbook two thirds if omitted) optional	—

You get

	Meaning	Unit	Example
\text{osm\_final}	Final osmolality in mOsm/kg	—	276.8181818
\text{ICF}	Final intracellular volume in L	—	29.33333333
\text{ECF}	Final extracellular volume in L	—	14.66666667

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

TBW, Posm, V_added > 0
C_added >= 0
solute stays in the ECF, water moves freely

Outside these conditions it says what is wrong instead of returning a number.

Worked example

fluid shift: TBW = 42 L, Posm = 290, V_added = 2 L, C_added = 0

osm_final = 276.8181818

ICF = 29.33333333

ECF = 14.66666667

Run it

Terminal

$ human-physiology solve hp_fluid_shift "TBW=42 L" Posm=290 "V_added=2 L" C_added=0
osm_final = 276.8181818
ICF = 29.33333333
ECF = 14.66666667
  by: Two-compartment fluid shift after adding a solution (Darrow-Yannet)  (solute conservation in each compartment, equal osmolality, water conservation)
  TBW = 0.042 m^3
  Posm = 290
  V_added = 0.002 m^3
  C_added = 0
  assumes: TBW, Posm, V_added > 0; C_added >= 0; solute stays in the ECF, water moves freely

import human_physiology_ai999x as m

a = m.solve("hp_fluid_shift", TBW="42 L", Posm="290", V_added="2 L", C_added="0")
print(a.values or (a.value, a.unit))

Figures

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Cardiovascular 26

Bernoulli pressure drop across a narrowing2 fig.hp_bernoulli_pressure_drop

Convective pressure drop 0.5 rho (v2^2 – v1^2) across a stenosis or valve, and the simplified 4 v^2 (mmHg, v in m/s).

\Delta P = \tfrac{1}{2}\rho\,(v_2^2 – v_1^2) \approx 4 v_2^2

You give

	Meaning	Unit	Example
\text{v1}	Upstream velocity (e.g. in m/s)	m/s	1 m/s
\text{v2}	Jet velocity (e.g. in m/s)	m/s	4 m/s
\rho	Blood density (e.g. in kg/m³)	kg/m^3	1060 kg/m^3

You get

	Meaning	Unit	Example
\text{dP}	Pressure drop in Pa	—	7950
\text{dP\_mmHg}	Pressure drop in mmHg	—	59.62989528
\text{dP\_simplified}	Pressure drop by the simplified Bernoulli equation in mmHg	—	64

Values with a unit may be written in any unit of the right kind.

Applies when

v2 > v1 >= 0
inviscid steady flow

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Bernoulli pressure drop: v1 = 1 m/s, v2 = 4 m/s, rho = 1060 kg/m^3

dP = 7950

dP_mmHg = 59.62989528

dP_simplified = 64

Run it

Terminal

$ human-physiology solve hp_bernoulli_pressure_drop "v1=1 m/s" "v2=4 m/s" "rho=1060 kg/m^3"
dP = 7950
dP_mmHg = 59.62989528
dP_simplified = 64
  by: Bernoulli pressure drop across a narrowing  (dP = rho (v2^2 - v1^2)/2)
  v1 = 1 m/s
  v2 = 4 m/s
  rho = 1060 kg/m^3
  assumes: v2 > v1 >= 0; inviscid steady flow

import human_physiology_ai999x as m

a = m.solve("hp_bernoulli_pressure_drop", v1="1 m/s", v2="4 m/s", rho="1060 kg/m^3")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Cardiac index and stroke volume index1 fig.hp_cardiac_index

Cardiac index CI = CO/BSA, stroke volume SV = CO/HR and stroke volume index SVI = CI/HR.

CI = \frac{CO}{BSA},\quad SVI = \frac{CI}{HR}

You give

	Meaning	Unit	Example
\text{CO}	Cardiac output (e.g. in L/min)	m^3/s	5.04 L/min
\text{BSA}	Body surface area (e.g. in m²)	m^2	1.8 m^2
\text{HR}	Heart rate in beats/min (optional; needed for the stroke volume index) optional	—	72

You get

	Meaning	Unit	Example
\text{CI}	Cardiac index in L/min/m²	—	2.8
\text{SVI}	Stroke volume index in mL/m²	—	38.88888889
\text{SV}	Stroke volume in mL	—	70

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

CO, BSA, HR > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

cardiac index: CO = 5.04 L/min, BSA = 1.8 m^2, HR = 72

CI = 2.8

SVI = 38.88888889

SV = 70

Run it

Terminal

$ human-physiology solve hp_cardiac_index "CO=5.04 L/min" "BSA=1.8 m^2" HR=72
CI = 2.8
SVI = 38.88888889
SV = 70
  by: Cardiac index and stroke volume index  (CI = CO/BSA; SVI = 1000 CI/HR)
  CO = 8.4e-05 m^3/s
  BSA = 1.8 m^2
  HR = 72
  assumes: CO, BSA, HR > 0

import human_physiology_ai999x as m

a = m.solve("hp_cardiac_index", CO="5.04 L/min", BSA="1.8 m^2", HR="72")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Cardiac output by the Fick principle1 fig.hp_fick_cardiac_output

Cardiac output from oxygen consumption and the arteriovenous oxygen content difference: CO = VO2/(CaO2 – CvO2).

CO = \frac{\dot V_{O_2}}{C_aO_2 – C_{\bar v}O_2}

You give

	Meaning	Unit	Example
\text{VO2}	Oxygen consumption (e.g. in mL/min)	m^3/s	250 mL/min
\text{CaO2}	Arterial oxygen content in mL O2/dL	—	20
\text{CvO2}	Mixed venous oxygen content in mL O2/dL	—	15

You get

	Meaning	Unit	Example
\text{CO}	Cardiac output in L/min	—	5
\text{avDO2}	Arteriovenous O2 content difference in mL/dL	—	5

Values with a unit may be written in any unit of the right kind.

Applies when

VO2 > 0
0 <= CvO2 < CaO2

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Fick cardiac output: VO2 = 250 mL/min, CaO2 = 20, CvO2 = 15

CO = 5

avDO2 = 5

Run it

Terminal

$ human-physiology solve hp_fick_cardiac_output "VO2=250 mL/min" CaO2=20 CvO2=15
CO = 5
avDO2 = 5
  by: Cardiac output by the Fick principle  (CO = VO2 / (10 (CaO2 - CvO2)) (contents per dL -> per L))
  VO2 = 4.166666667e-06 m^3/s
  CaO2 = 20
  CvO2 = 15
  assumes: VO2 > 0; 0 <= CvO2 < CaO2

import human_physiology_ai999x as m

a = m.solve("hp_fick_cardiac_output", VO2="250 mL/min", CaO2="20", CvO2="15")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Cardiac output from heart rate and stroke volume2 fig.hp_cardiac_output

Cardiac output CO = HR x SV, in L/min and mL/min.

CO = HR \times SV

You give

	Meaning	Unit	Example
\text{HR}	Heart rate in beats/min	—	72
\text{SV}	Stroke volume (e.g. in mL)	m^3	70 mL

You get

	Meaning	Unit	Example
\text{CO}	Cardiac output in L/min	—	5.04
\text{CO\_mL}	Cardiac output in mL/min	—	5040

Values with a unit may be written in any unit of the right kind.

Applies when

HR > 0
SV > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

cardiac output: HR = 72, SV = 70 mL

CO = 5.04

CO_mL = 5040

Run it

Terminal

$ human-physiology solve hp_cardiac_output HR=72 "SV=70 mL"
CO = 5.04
CO_mL = 5040
  by: Cardiac output from heart rate and stroke volume  (CO = HR SV / 1000)
  HR = 72
  SV = 7e-05 m^3
  assumes: HR > 0; SV > 0

import human_physiology_ai999x as m

a = m.solve("hp_cardiac_output", HR="72", SV="70 mL")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Compliance and elastance2 fig.hp_vascular_compliance

Compliance C = dV/dP and its inverse, the elastance E = dP/dV.

C = \frac{\Delta V}{\Delta P}

You give

	Meaning	Unit	Example
\text{dV}	Volume change (e.g. in mL)	m^3	70 mL
\text{dP}	Pressure change (e.g. in mmHg)	Pa	40 mmHg

You get

	Meaning	Unit	Example
C	Compliance in mL/mmHg	—	1.75
E	Elastance in mmHg/mL	—	0.5714285714

Values with a unit may be written in any unit of the right kind.

Applies when

dV, dP > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

vascular compliance: dV = 70 mL, dP = 40 mmHg

C = 1.75

E = 0.5714285714

Run it

Terminal

$ human-physiology solve hp_vascular_compliance "dV=70 mL" "dP=40 mmHg"
C = 1.75
E = 0.5714285714
  by: Compliance and elastance  (C = dV/dP)
  dV = 7e-05 m^3
  dP = 5332.895497 Pa
  assumes: dV, dP > 0

import human_physiology_ai999x as m

a = m.solve("hp_vascular_compliance", dV="70 mL", dP="40 mmHg")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Effect of a change of vessel radius on flow and resistance2 fig.hp_radius_flow_ratio

At the same pressure drop, flow scales as r^4 and resistance as 1/r^4 (Poiseuille).

\frac{Q_2}{Q_1} = \left(\frac{r_2}{r_1}\right)^4

You give

	Meaning	Unit	Example
\text{r1}	Initial radius (any unit)	—	1
\text{r2}	New radius (same unit)	—	2

You get

	Meaning	Unit	Example
\text{flow\_ratio}	New flow / initial flow	—	16
\text{resistance\_ratio}	New resistance / initial resistance	—	0.0625

Applies when

r1, r2 > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

radius change and flow: r1 = 1, r2 = 2

flow_ratio = 16

resistance_ratio = 0.0625

Run it

Terminal

$ human-physiology solve hp_radius_flow_ratio r1=1 r2=2
flow_ratio = 16
resistance_ratio = 0.0625
  by: Effect of a change of vessel radius on flow and resistance  (Q2/Q1 = (r2/r1)^4)
  r1 = 1
  r2 = 2
  assumes: r1, r2 > 0

import human_physiology_ai999x as m

a = m.solve("hp_radius_flow_ratio", r1="1", r2="2")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Fick cardiac output from hemoglobin, saturations and PO21 fig.hp_fick_saturations

Arterial and mixed venous O2 contents (1.34 Hb S + 0.003 PO2) and the Fick cardiac output VO2/(CaO2 – CvO2).

CO = \frac{\dot V_{O_2}}{1.34\,Hb\,(S_aO_2 – S_{\bar v}O_2) + 0.003\,(P_aO_2 – P_{\bar v}O_2)}

You give

	Meaning	Unit	Example
\text{VO2}	Oxygen consumption (e.g. in mL/min)	m^3/s	250 mL/min
\text{Hb}	Hemoglobin in g/dL	—	15
\text{SaO2}	Arterial saturation (fraction)	—	0.98
\text{SvO2}	Mixed venous saturation (fraction)	—	0.75
\text{PaO2}	Arterial PO2 (e.g. in mmHg)	Pa	95 mmHg
\text{PvO2}	Mixed venous PO2 (e.g. in mmHg)	Pa	40 mmHg

You get

	Meaning	Unit	Example
\text{CO}	Cardiac output in L/min	—	5.2213868
\text{CaO2}	Arterial O2 content in mL/dL	—	19.983
\text{CvO2}	Venous O2 content in mL/dL	—	15.195

Values with a unit may be written in any unit of the right kind.

Applies when

saturations between 0 and 1
CvO2 < CaO2

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Fick cardiac output from saturations: VO2 = 250 mL/min, Hb = 15, SaO2 = 0.98, SvO2 = 0.75, PaO2 = 95 mmHg, PvO2 = 40 mmHg

CO = 5.2213868

CaO2 = 19.983

CvO2 = 15.195

Run it

Terminal

$ human-physiology solve hp_fick_saturations "VO2=250 mL/min" Hb=15 SaO2=0.98 SvO2=0.75 "PaO2=95 mmHg" "PvO2=40 mmHg"
CO = 5.2213868
CaO2 = 19.983
CvO2 = 15.195
  by: Fick cardiac output from hemoglobin, saturations and PO2  (C = 1.34 Hb S + 0.003 PO2; CO = VO2/(10 (CaO2 - CvO2)))
  VO2 = 4.166666667e-06 m^3/s
  Hb = 15
  SaO2 = 0.98
  SvO2 = 0.75
  PaO2 = 12665.6268 Pa
  PvO2 = 5332.895497 Pa
  assumes: saturations between 0 and 1; CvO2 < CaO2

import human_physiology_ai999x as m

a = m.solve("hp_fick_saturations", VO2="250 mL/min", Hb="15", SaO2="0.98", SvO2="0.75", PaO2="95 mmHg", PvO2="40 mmHg")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Heart-rate corrected QT interval (Bazett, Fridericia, Framingham)2 fig.hp_qt_correction

RR = 60/HR and the corrected QT: Bazett QT/sqrt(RR), Fridericia QT/RR^(1/3), Framingham QT + 154 (1 – RR).

QT_c = \frac{QT}{\sqrt{RR}} \;(\text{Bazett}),\quad \frac{QT}{RR^{1/3}} \;(\text{Fridericia})

You give

	Meaning	Unit	Example
\text{QT}	Measured QT interval (e.g. in ms)	s	400 ms
\text{HR}	Heart rate in beats/min	—	75

You get

	Meaning	Unit	Example
\text{RR}	RR interval in s	—	0.8
\text{QTc\_Bazett}	Bazett QTc in ms	—	447.2135955
\text{QTc\_Fridericia}	Fridericia QTc in ms	—	430.886938
\text{QTc\_Framingham}	Framingham QTc in ms	—	430.8

Values with a unit may be written in any unit of the right kind.

Applies when

QT, HR > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

corrected QT interval: QT = 400 ms, HR = 75

RR = 0.8

QTc_Bazett = 447.2135955

QTc_Fridericia = 430.886938

QTc_Framingham = 430.8

Run it

Terminal

$ human-physiology solve hp_qt_correction "QT=400 ms" HR=75
RR = 0.8
QTc_Bazett = 447.2135955
QTc_Fridericia = 430.886938
QTc_Framingham = 430.8
  by: Heart-rate corrected QT interval (Bazett, Fridericia, Framingham)  (RR in seconds; the three published corrections)
  QT = 0.4 s
  HR = 75
  assumes: QT, HR > 0

import human_physiology_ai999x as m

a = m.solve("hp_qt_correction", QT="400 ms", HR="75")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Hydrostatic pressure of a blood column2 fig.hp_hydrostatic_pressure

Pressure difference rho g h between two points a height h apart in a column of blood (posture effects).

\Delta P = \rho g h

You give

	Meaning	Unit	Example
h	Height of the column (e.g. in cm) (positive = the lower point)	m	130 cm
\rho	Blood density (e.g. in kg/m³)	kg/m^3	1060 kg/m^3

You get

	Meaning	Unit	Example
\text{dP}	Pressure difference in Pa	—	13513.5637
\text{dP\_mmHg}	Pressure difference in mmHg	—	101.3600488

Values with a unit may be written in any unit of the right kind.

Applies when

rho > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

hydrostatic pressure of a blood column: h = 130 cm, rho = 1060 kg/m^3

dP = 13513.5637

dP_mmHg = 101.3600488

Run it

Terminal

$ human-physiology solve hp_hydrostatic_pressure "h=130 cm" "rho=1060 kg/m^3"
dP = 13513.5637
dP_mmHg = 101.3600488
  by: Hydrostatic pressure of a blood column  (dP = rho g h)
  h = 1.3 m
  rho = 1060 kg/m^3
  assumes: rho > 0

import human_physiology_ai999x as m

a = m.solve("hp_hydrostatic_pressure", h="130 cm", rho="1060 kg/m^3")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Laplace wall tension and wall stress of a vessel or chamber1 fig.hp_laplace_wall_tension

Law of Laplace: wall tension T = P r (cylinder) or P r/2 (sphere) and wall stress T/h.

T = P r \;(\text{cylinder}),\quad T = \tfrac{P r}{2} \;(\text{sphere}),\quad \sigma = T/h

You give

	Meaning	Unit	Example
P	Transmural pressure (e.g. in mmHg)	Pa	100 mmHg
r	Radius (e.g. in cm)	m	1.5 cm
\text{shape}	Cylinder (vessel) or sphere (chamber, alveolus) choice	—	cylinder
h	Wall thickness (e.g. in mm) (optional; needed for the wall stress) optional	m	2 mm

You get

	Meaning	Unit	Example
T	Wall tension in N/m	—	199.9835811
\text{stress}	Wall stress in kPa	—	99.99179056

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

P, r, h > 0
thin wall

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Laplace wall tension: P = 100 mmHg, r = 1.5 cm, shape = cylinder, h = 2 mm

T = 199.9835811

stress = 99.99179056

Run it

Terminal

$ human-physiology solve hp_laplace_wall_tension "P=100 mmHg" "r=1.5 cm" shape=cylinder "h=2 mm"
T = 199.9835811
stress = 99.99179056
  by: Laplace wall tension and wall stress of a vessel or chamber  (T = P r (cylinder), P r/2 (sphere); stress = T/h)
  P = 13332.23874 Pa
  r = 0.015 m
  h = 0.002 m
  shape = cylinder
  assumes: P, r, h > 0; thin wall

import human_physiology_ai999x as m

a = m.solve("hp_laplace_wall_tension", P="100 mmHg", r="1.5 cm", shape="cylinder", h="2 mm")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Left-ventricular volumes and ejection fraction by the Teichholz formula2 fig.hp_teichholz_volumes

Teichholz volume V = 7 D^3/(2.4 + D) from M-mode internal diameters, EF and fractional shortening.

V = \frac{7 D^3}{2.4 + D}

You give

	Meaning	Unit	Example
\text{LVIDd}	End-diastolic internal diameter (e.g. in cm)	m	5 cm
\text{LVIDs}	End-systolic internal diameter (e.g. in cm)	m	3.2 cm

You get

	Meaning	Unit	Example
\text{EDV}	End-diastolic volume in mL	—	118.2432432
\text{ESV}	End-systolic volume in mL	—	40.96
\text{EF}	Ejection fraction	—	0.6535954286
\text{FS}	Fractional shortening	—	0.36

Values with a unit may be written in any unit of the right kind.

Applies when

0 < LVIDs < LVIDd

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Teichholz volumes: LVIDd = 5 cm, LVIDs = 3.2 cm

EDV = 118.2432432

ESV = 40.96

EF = 0.6535954286

FS = 0.36

Run it

Terminal

$ human-physiology solve hp_teichholz_volumes "LVIDd=5 cm" "LVIDs=3.2 cm"
EDV = 118.2432432
ESV = 40.96
EF = 0.6535954286
FS = 0.36
  by: Left-ventricular volumes and ejection fraction by the Teichholz formula  (V = 7 D^3/(2.4 + D) (D in cm, V in mL))
  LVIDd = 0.05 m
  LVIDs = 0.032 m
  assumes: 0 < LVIDs < LVIDd

import human_physiology_ai999x as m

a = m.solve("hp_teichholz_volumes", LVIDd="5 cm", LVIDs="3.2 cm")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Mean arterial pressure and pulse pressure2 fig.hp_mean_arterial_pressure

Pulse pressure PP = SBP – DBP and the estimated mean arterial pressure MAP = DBP + PP/3.

MAP = DBP + \tfrac{1}{3}(SBP – DBP)

You give

	Meaning	Unit	Example
\text{SBP}	Systolic pressure (e.g. in mmHg)	Pa	120 mmHg
\text{DBP}	Diastolic pressure (e.g. in mmHg)	Pa	80 mmHg

You get

	Meaning	Unit	Example
\text{MAP}	Mean arterial pressure in mmHg	—	93.33333333
\text{PP}	Pulse pressure in mmHg	—	40

Values with a unit may be written in any unit of the right kind.

Applies when

SBP > DBP > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

mean arterial pressure: SBP = 120 mmHg, DBP = 80 mmHg

MAP = 93.33333333

PP = 40

Run it

Terminal

$ human-physiology solve hp_mean_arterial_pressure "SBP=120 mmHg" "DBP=80 mmHg"
MAP = 93.33333333
PP = 40
  by: Mean arterial pressure and pulse pressure  (MAP = DBP + (SBP - DBP)/3)
  SBP = 15998.68649 Pa
  DBP = 10665.79099 Pa
  assumes: SBP > DBP > 0

import human_physiology_ai999x as m

a = m.solve("hp_mean_arterial_pressure", SBP="120 mmHg", DBP="80 mmHg")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Mean blood velocity from flow and total cross-sectional area2 fig.hp_blood_velocity_area

Continuity: mean velocity v = Q/A (slow in the capillaries because their total area is large).

v = \frac{Q}{A}

You give

	Meaning	Unit	Example
Q	Flow (e.g. in L/min)	m^3/s	5 L/min
A	Total cross-sectional area (e.g. in cm²)	m^2	4 cm^2

You get

	Meaning	Unit	Example
v	Mean velocity in cm/s	—	20.83333333

Values with a unit may be written in any unit of the right kind.

Applies when

Q, A > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

mean blood velocity: Q = 5 L/min, A = 4 cm^2

v = 20.83333333

Run it

Terminal

$ human-physiology solve hp_blood_velocity_area "Q=5 L/min" "A=4 cm^2"
v = 20.83333333
  by: Mean blood velocity from flow and total cross-sectional area  (v = Q/A (L/min -> cm^3/s))
  Q = 8.333333333e-05 m^3/s
  A = 0.0004 m^2
  assumes: Q, A > 0

import human_physiology_ai999x as m

a = m.solve("hp_blood_velocity_area", Q="5 L/min", A="4 cm^2")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Mean QRS (electrical) axis from leads I and aVF2 fig.hp_qrs_axis

The frontal-plane mean electrical axis atan2(aVF, I) in degrees (hexaxial convention: lead I at 0 deg, aVF at +90 deg).

\theta = \operatorname{atan2}(aVF, I)

You give

	Meaning	Unit	Example
\text{lead\_I}	Net QRS deflection in lead I (mm or mV)	—	0.5
\text{aVF}	Net QRS deflection in aVF (same unit)	—	-0.5

You get

	Meaning	Unit	Example
\text{axis}	Mean QRS axis in degrees (principal value)	—	-45
\text{amplitude}	Length of the mean vector (same unit)	—	0.7071067812

Applies when

not both zero

Outside these conditions it says what is wrong instead of returning a number.

Worked example

QRS axis: lead_I = 0.5, aVF = -0.5

axis = -45

amplitude = 0.7071067812

Run it

Terminal

$ human-physiology solve hp_qrs_axis lead_I=0.5 aVF=-0.5
axis = -45
amplitude = 0.7071067812
  by: Mean QRS (electrical) axis from leads I and aVF  (axis = atan2(aVF, I))
  lead_I = 0.5
  aVF = -0.5
  assumes: not both zero

import human_physiology_ai999x as m

a = m.solve("hp_qrs_axis", lead_I="0.5", aVF="-0.5")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Poiseuille flow, mean velocity and wall shear stress in a vessel1 fig.hp_poiseuille_flow

Laminar flow Q = dP pi r^4/(8 eta L) through a vessel, its mean velocity and the wall shear stress 4 eta Q/(pi r^3).

Q = \frac{\pi r^4 \Delta P}{8 \eta L},\quad \tau_w = \frac{4\eta Q}{\pi r^3}

You give

	Meaning	Unit	Example
\text{dP}	Pressure drop along the vessel (e.g. in mmHg)	Pa	10 mmHg
\eta	Blood viscosity in cP (mPa s)	—	3.5
L	Vessel length (e.g. in cm)	m	10 cm
r	Vessel radius (e.g. in mm)	m	2 mm

You get

	Meaning	Unit	Example
Q	Flow in mL/s	—	23.93397902
Q_{\mathrm{mL},\mathrm{min}}	Flow in mL/min	—	1436.038741
v_{\mathrm{mean}}	Mean velocity in cm/s	—	190.4605534
\tau_{\mathrm{wall}}	Wall shear stress in Pa	—	13.33223874

Values with a unit may be written in any unit of the right kind.

Applies when

all inputs > 0
steady laminar Newtonian flow

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Poiseuille flow: dP = 10 mmHg, eta = 3.5, L = 10 cm, r = 2 mm

Q = 23.93397902

Q_mL_min = 1436.038741

v_mean = 190.4605534

tau_wall = 13.33223874

Run it

Terminal

$ human-physiology solve hp_poiseuille_flow "dP=10 mmHg" eta=3.5 "L=10 cm" "r=2 mm"
Q = 23.93397902
Q_mL_min = 1436.038741
v_mean = 190.4605534
tau_wall = 13.33223874
  by: Poiseuille flow, mean velocity and wall shear stress in a vessel  (Q = dP pi r^4 /(8 eta L); tau_w = 4 eta Q/(pi r^3))
  dP = 1333.223874 Pa
  eta = 3.5
  L = 0.1 m
  r = 0.002 m
  assumes: all inputs > 0; steady laminar Newtonian flow

import human_physiology_ai999x as m

a = m.solve("hp_poiseuille_flow", dP="10 mmHg", eta="3.5", L="10 cm", r="2 mm")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Poiseuille resistance of a blood vessel1 fig.hp_poiseuille_resistance

Resistance to laminar flow of a cylindrical vessel R = 8 eta L/(pi r^4), in Pa s/m³ and in mmHg min/mL.

R = \frac{8\eta L}{\pi r^4}

You give

	Meaning	Unit	Example
\eta	Blood viscosity in cP (mPa s)	—	3.5
L	Vessel length (e.g. in cm)	m	10 cm
r	Vessel radius (e.g. in mm)	m	2 mm

You get

	Meaning	Unit	Example
R	Resistance in Pa s/m³	—	55704230.08
R_{\mathrm{mmHg}}	Resistance in mmHg min/mL	—	0.006963600433

Values with a unit may be written in any unit of the right kind.

Applies when

eta, L, r > 0
steady laminar Newtonian flow in a rigid tube

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Poiseuille resistance: eta = 3.5, L = 10 cm, r = 2 mm

R = 55704230.08

R_mmHg = 0.006963600433

Run it

Terminal

$ human-physiology solve hp_poiseuille_resistance eta=3.5 "L=10 cm" "r=2 mm"
R = 55704230.08
R_mmHg = 0.006963600433
  by: Poiseuille resistance of a blood vessel  (R = 8 eta L/(pi r^4))
  eta = 3.5
  L = 0.1 m
  r = 0.002 m
  assumes: eta, L, r > 0; steady laminar Newtonian flow in a rigid tube

import human_physiology_ai999x as m

a = m.solve("hp_poiseuille_resistance", eta="3.5", L="10 cm", r="2 mm")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Pulmonary vascular resistance and transpulmonary gradient1 fig.hp_pulmonary_vascular_resistance

Transpulmonary gradient TPG = mPAP – PCWP and PVR = TPG/CO in Wood units and dyn s cm^-5.

PVR = \frac{\bar P_{PA} – P_{CW}}{CO}

You give

	Meaning	Unit	Example
\text{mPAP}	Mean pulmonary artery pressure (e.g. in mmHg)	Pa	15 mmHg
\text{PCWP}	Pulmonary capillary wedge pressure (e.g. in mmHg)	Pa	8 mmHg
\text{CO}	Cardiac output (e.g. in L/min)	m^3/s	5 L/min

You get

	Meaning	Unit	Example
\text{PVR}	Pulmonary vascular resistance in Wood units	—	1.4
\text{PVR\_dyn}	Pulmonary vascular resistance in dyn s cm^-5	—	111.9908054
\text{TPG}	Transpulmonary gradient in mmHg	—	7

Values with a unit may be written in any unit of the right kind.

Applies when

mPAP > PCWP >= 0
CO > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

pulmonary vascular resistance: mPAP = 15 mmHg, PCWP = 8 mmHg, CO = 5 L/min

PVR = 1.4

PVR_dyn = 111.9908054

TPG = 7

Run it

Terminal

$ human-physiology solve hp_pulmonary_vascular_resistance "mPAP=15 mmHg" "PCWP=8 mmHg" "CO=5 L/min"
PVR = 1.4
PVR_dyn = 111.9908054
TPG = 7
  by: Pulmonary vascular resistance and transpulmonary gradient  (PVR = (mPAP - PCWP)/CO)
  mPAP = 1999.835811 Pa
  PCWP = 1066.579099 Pa
  CO = 8.333333333e-05 m^3/s
  assumes: mPAP > PCWP >= 0; CO > 0

import human_physiology_ai999x as m

a = m.solve("hp_pulmonary_vascular_resistance", mPAP="15 mmHg", PCWP="8 mmHg", CO="5 L/min")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Pulmonary-to-systemic flow ratio Qp/Qs1 fig.hp_qp_qs

Qp/Qs = (SaO2 – SvO2)/(SpvO2 – SpaO2) from oxygen saturations (hemoglobin and dissolved O2 cancel or are neglected).

\frac{Q_p}{Q_s} = \frac{S_aO_2 – S_{\bar v}O_2}{S_{pv}O_2 – S_{pa}O_2}

You give

	Meaning	Unit	Example
\text{SaO2}	Systemic arterial saturation	—	0.95
\text{SvO2}	Mixed venous saturation	—	0.65
\text{SpvO2}	Pulmonary venous saturation	—	0.98
\text{SpaO2}	Pulmonary artery saturation	—	0.8

You get

	Meaning	Unit	Example
\text{Qp\_Qs}	Pulmonary to systemic flow ratio	—	1.666666667

Applies when

SaO2 > SvO2
SpvO2 > SpaO2

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Qp/Qs: SaO2 = 0.95, SvO2 = 0.65, SpvO2 = 0.98, SpaO2 = 0.8

Qp_Qs = 1.666666667

Run it

Terminal

$ human-physiology solve hp_qp_qs SaO2=0.95 SvO2=0.65 SpvO2=0.98 SpaO2=0.8
Qp_Qs = 1.666666667
  by: Pulmonary-to-systemic flow ratio Qp/Qs  (Fick principle applied to both circulations)
  SaO2 = 0.95
  SvO2 = 0.65
  SpvO2 = 0.98
  SpaO2 = 0.8
  assumes: SaO2 > SvO2; SpvO2 > SpaO2

import human_physiology_ai999x as m

a = m.solve("hp_qp_qs", SaO2="0.95", SvO2="0.65", SpvO2="0.98", SpaO2="0.8")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Pulse wave velocity by the Moens-Korteweg equation1 fig.hp_moens_korteweg

Pulse wave velocity of a thin elastic tube PWV = sqrt(E h/(rho D)).

PWV = \sqrt{\frac{E h}{\rho D}}

You give

	Meaning	Unit	Example
E	Wall elastic modulus (e.g. in kPa)	Pa	500 kPa
h	Wall thickness (e.g. in mm)	m	1.5 mm
D	Vessel diameter (e.g. in mm)	m	25 mm
\rho	Blood density (e.g. in kg/m³)	kg/m^3	1060 kg/m^3

You get

	Meaning	Unit	Example
\text{PWV}	Pulse wave velocity in m/s	—	5.319951766

Values with a unit may be written in any unit of the right kind.

Applies when

all inputs > 0
thin-walled elastic tube

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Moens-Korteweg pulse wave velocity: E = 500 kPa, h = 1.5 mm, D = 25 mm, rho = 1060 kg/m^3

PWV = 5.319951766

Run it

Terminal

$ human-physiology solve hp_moens_korteweg "E=500 kPa" "h=1.5 mm" "D=25 mm" "rho=1060 kg/m^3"
PWV = 5.319951766
  by: Pulse wave velocity by the Moens-Korteweg equation  (PWV = sqrt(E h /(rho D)))
  E = 500000 Pa
  h = 0.0015 m
  D = 0.025 m
  rho = 1060 kg/m^3
  assumes: all inputs > 0; thin-walled elastic tube

import human_physiology_ai999x as m

a = m.solve("hp_moens_korteweg", E="500 kPa", h="1.5 mm", D="25 mm", rho="1060 kg/m^3")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Reynolds number of blood flow in a vessel1 fig.hp_vessel_reynolds

Re = rho v D/eta and the mean velocity at which Re reaches 2000 (the customary laminar limit).

Re = \frac{\rho v D}{\eta}

You give

	Meaning	Unit	Example
\rho	Blood density (e.g. in kg/m³)	kg/m^3	1060 kg/m^3
v	Mean velocity (e.g. in cm/s)	m/s	30 cm/s
D	Vessel diameter (e.g. in mm)	m	25 mm
\eta	Viscosity in cP	—	3.5

You get

	Meaning	Unit	Example
\text{Re}	Reynolds number	—	2271.428571
v_{\mathrm{critical}}	Mean velocity at the customary laminar-limit Reynolds number in cm/s	—	26.41509434

Values with a unit may be written in any unit of the right kind.

Applies when

all inputs > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

vessel Reynolds number: rho = 1060 kg/m^3, v = 30 cm/s, D = 25 mm, eta = 3.5

Re = 2271.428571

v_critical = 26.41509434

Run it

Terminal

$ human-physiology solve hp_vessel_reynolds "rho=1060 kg/m^3" "v=30 cm/s" "D=25 mm" eta=3.5
Re = 2271.428571
v_critical = 26.41509434
  by: Reynolds number of blood flow in a vessel  (Re = rho v D / eta)
  rho = 1060 kg/m^3
  v = 0.3 m/s
  D = 0.025 m
  eta = 3.5
  assumes: all inputs > 0

import human_physiology_ai999x as m

a = m.solve("hp_vessel_reynolds", rho="1060 kg/m^3", v="30 cm/s", D="25 mm", eta="3.5")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Starling forces and capillary filtration1 fig.hp_starling_filtration

Net filtration pressure (Pc – Pi) – sigma (pi_c – pi_i) and the filtration rate Jv = Kf x NFP (glomerular or systemic capillaries).

J_v = K_f\left[(P_c – P_i) – \sigma(\pi_c – \pi_i)\right]

You give

	Meaning	Unit	Example
\text{Pc}	Capillary hydrostatic pressure (e.g. in mmHg)	Pa	35 mmHg
\Pi	Interstitial (or Bowman's space) hydrostatic pressure (e.g. in mmHg)	Pa	0 mmHg
\text{pic}	Capillary oncotic pressure (e.g. in mmHg)	Pa	28 mmHg
\text{pii}	Interstitial oncotic pressure (e.g. in mmHg)	Pa	3 mmHg
\text{Kf}	Filtration coefficient in mL/min per mmHg	—	0.5
\sigma	Reflection coefficient (optional; ideal membrane if omitted) optional	—

You get

	Meaning	Unit	Example
\text{NFP}	Net filtration pressure in mmHg (negative = absorption)	—	10
\text{Jv}	Filtration rate in mL/min	—	5

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

Kf > 0
0 <= sigma <= 1

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Starling filtration: Pc = 35 mmHg, Pi = 0 mmHg, pic = 28 mmHg, pii = 3 mmHg, Kf = 0.5

NFP = 10

Jv = 5

Run it

Terminal

$ human-physiology solve hp_starling_filtration "Pc=35 mmHg" "Pi=0 mmHg" "pic=28 mmHg" "pii=3 mmHg" Kf=0.5
NFP = 10
Jv = 5
  by: Starling forces and capillary filtration  (Jv = Kf [(Pc - Pi) - sigma (pi_c - pi_i)])
  Pc = 4666.28356 Pa
  Pi = 0 Pa
  pic = 3733.026848 Pa
  pii = 399.9671622 Pa
  Kf = 0.5
  assumes: Kf > 0; 0 <= sigma <= 1

import human_physiology_ai999x as m

a = m.solve("hp_starling_filtration", Pc="35 mmHg", Pi="0 mmHg", pic="28 mmHg", pii="3 mmHg", Kf="0.5")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Stroke volume and ejection fraction2 fig.hp_stroke_volume_ef

Stroke volume SV = EDV – ESV and ejection fraction EF = SV/EDV from end-diastolic and end-systolic volumes.

SV = EDV – ESV,\quad EF = \frac{SV}{EDV}

You give

	Meaning	Unit	Example
\text{EDV}	End-diastolic volume (e.g. in mL)	m^3	120 mL
\text{ESV}	End-systolic volume (e.g. in mL)	m^3	50 mL

You get

	Meaning	Unit	Example
\text{SV}	Stroke volume in mL	—	70
\text{EF}	Ejection fraction (fraction)	—	0.5833333333
\text{EF\_percent}	Ejection fraction in %	—	58.33333333

Values with a unit may be written in any unit of the right kind.

Applies when

0 <= ESV < EDV

Outside these conditions it says what is wrong instead of returning a number.

Worked example

stroke volume and ejection fraction: EDV = 120 mL, ESV = 50 mL

SV = 70

EF = 0.5833333333

EF_percent = 58.33333333

Run it

Terminal

$ human-physiology solve hp_stroke_volume_ef "EDV=120 mL" "ESV=50 mL"
SV = 70
EF = 0.5833333333
EF_percent = 58.33333333
  by: Stroke volume and ejection fraction  (SV = EDV - ESV; EF = SV/EDV)
  EDV = 0.00012 m^3
  ESV = 5e-05 m^3
  assumes: 0 <= ESV < EDV

import human_physiology_ai999x as m

a = m.solve("hp_stroke_volume_ef", EDV="120 mL", ESV="50 mL")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Stroke work and cardiac power1 fig.hp_stroke_work

External stroke work SW = MAP x SV (J and g m) and the cardiac power output SW x HR.

SW = \bar P\, SV,\quad \dot W = SW \cdot HR

You give

	Meaning	Unit	Example
\text{SV}	Stroke volume (e.g. in mL)	m^3	70 mL
\text{MAP}	Mean arterial (ejection) pressure (e.g. in mmHg)	Pa	100 mmHg
\text{HR}	Heart rate in beats/min (optional; needed for the cardiac power) optional	—	72

You get

	Meaning	Unit	Example
\text{SW}	Stroke work in J	—	0.9332567119
\text{SW\_gm}	Stroke work in g m	—	95.1657
\text{power}	Cardiac power output in W	—	1.119908054

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

SV, MAP, HR > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

stroke work: SV = 70 mL, MAP = 100 mmHg, HR = 72

SW = 0.9332567119

SW_gm = 95.1657

power = 1.119908054

Run it

Terminal

$ human-physiology solve hp_stroke_work "SV=70 mL" "MAP=100 mmHg" HR=72
SW = 0.9332567119
SW_gm = 95.1657
power = 1.119908054
  by: Stroke work and cardiac power  (SW = MAP SV (pressure-volume work); power = SW HR/60)
  SV = 7e-05 m^3
  MAP = 13332.23874 Pa
  HR = 72
  assumes: SV, MAP, HR > 0

import human_physiology_ai999x as m

a = m.solve("hp_stroke_work", SV="70 mL", MAP="100 mmHg", HR="72")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Systemic vascular resistance (total peripheral resistance)1 fig.hp_systemic_vascular_resistance

SVR = (MAP – CVP)/CO in Wood units (mmHg min/L) and in dyn s cm^-5 (exact factor 79.9934; customary 80).

SVR = \frac{MAP – CVP}{CO}

You give

	Meaning	Unit	Example
\text{MAP}	Mean arterial pressure (e.g. in mmHg)	Pa	93 mmHg
\text{CVP}	Central venous pressure (e.g. in mmHg)	Pa	3 mmHg
\text{CO}	Cardiac output (e.g. in L/min)	m^3/s	5 L/min

You get

	Meaning	Unit	Example
\text{SVR}	Systemic vascular resistance in mmHg min/L (Wood units)	—	18
\text{SVR\_dyn}	Systemic vascular resistance in dyn s cm^-5	—	1439.881784

Values with a unit may be written in any unit of the right kind.

Applies when

MAP > CVP >= 0
CO > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

systemic vascular resistance: MAP = 93 mmHg, CVP = 3 mmHg, CO = 5 L/min

SVR = 18

SVR_dyn = 1439.881784

Run it

Terminal

$ human-physiology solve hp_systemic_vascular_resistance "MAP=93 mmHg" "CVP=3 mmHg" "CO=5 L/min"
SVR = 18
SVR_dyn = 1439.881784
  by: Systemic vascular resistance (total peripheral resistance)  (SVR = (MAP - CVP)/CO; x 79.9934 for dyn s cm^-5)
  MAP = 12398.98203 Pa
  CVP = 399.9671622 Pa
  CO = 8.333333333e-05 m^3/s
  assumes: MAP > CVP >= 0; CO > 0

import human_physiology_ai999x as m

a = m.solve("hp_systemic_vascular_resistance", MAP="93 mmHg", CVP="3 mmHg", CO="5 L/min")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Total resistance of vascular beds in series or in parallel1 fig.hp_vessels_series_parallel

Series beds add resistances; parallel beds (organ circulations) add conductances.

R_s = \sum R_i,\quad \frac{1}{R_p} = \sum \frac{1}{R_i}

You give

	Meaning	Unit	Example
R	Resistances of the beds (any one unit, e.g. mmHg min/L) vector	—	[2, 3, 6]
\text{arrangement}	Series or parallel choice	—	parallel

You get

	Meaning	Unit	Example
R_{\mathrm{total}}	Total resistance (same unit)	—	1
G_{\mathrm{total}}	Total conductance (reciprocal of the resistance unit)	—	1

Applies when

every resistance > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

vessels in series or parallel: R = [2, 3, 6], arrangement = parallel

R_total = 1

G_total = 1

Run it

Terminal

$ human-physiology solve hp_vessels_series_parallel "R=[2, 3, 6]" arrangement=parallel
R_total = 1
G_total = 1
  by: Total resistance of vascular beds in series or in parallel  (series: sum R; parallel: 1/sum(1/R))
  arrangement = parallel
  assumes: every resistance > 0

import human_physiology_ai999x as m

a = m.solve("hp_vessels_series_parallel", R=[2, 3, 6], arrangement="parallel")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Womersley number of pulsatile blood flow1 fig.hp_womersley_number

alpha = r sqrt(omega rho/eta) with omega = 2 pi HR/60: the ratio of pulsatile inertia to viscous forces.

\alpha = r\sqrt{\frac{\omega\rho}{\eta}}

You give

	Meaning	Unit	Example
r	Vessel radius (e.g. in mm)	m	12.5 mm
\text{HR}	Heart rate in beats/min	—	72
\rho	Blood density (e.g. in kg/m³)	kg/m^3	1060 kg/m^3
\eta	Viscosity in cP	—	3.5

You get

	Meaning	Unit	Example
\alpha	Womersley number	—	18.88902236
\omega	Angular frequency in rad/s	—	7.539822369

Values with a unit may be written in any unit of the right kind.

Applies when

all inputs > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Womersley number: r = 12.5 mm, HR = 72, rho = 1060 kg/m^3, eta = 3.5

alpha = 18.88902236

omega = 7.539822369

Run it

Terminal

$ human-physiology solve hp_womersley_number "r=12.5 mm" HR=72 "rho=1060 kg/m^3" eta=3.5
alpha = 18.88902236
omega = 7.539822369
  by: Womersley number of pulsatile blood flow  (alpha = r sqrt(omega rho / eta))
  r = 0.0125 m
  HR = 72
  rho = 1060 kg/m^3
  eta = 3.5
  assumes: all inputs > 0

import human_physiology_ai999x as m

a = m.solve("hp_womersley_number", r="12.5 mm", HR="72", rho="1060 kg/m^3", eta="3.5")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Metabolism 8

Basal metabolic rate by the Harris-Benedict equation1 fig.hp_bmr_harris_benedict

Harris-Benedict BMR, original (1919) or revised by Roza and Shizgal (1984), in kcal/day and kJ/day (computed value).

BMR_{m} = 66.473 + 13.7516\,W + 5.0033\,H – 6.755\,A

You give

	Meaning	Unit	Example
\text{weight}	Weight (e.g. in kg)	kg	70 kg
\text{height}	Height (e.g. in cm)	m	175 cm
\text{age}	Age in years	—	30
\text{sex}	Sex for the coefficients choice	—	male
\text{version}	Original equation or the revised (Roza-Shizgal) equation choice	—	original

You get

	Meaning	Unit	Example
\text{BMR}	Basal metabolic rate in kcal/day	—	1702.0125
\text{BMR\_kJ}	Basal metabolic rate in kJ/day	—	7121.2203

Values with a unit may be written in any unit of the right kind.

Applies when

adult
weight, height > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Harris-Benedict BMR: weight = 70 kg, height = 175 cm, age = 30, sex = male, version = original

BMR = 1702.0125

BMR_kJ = 7121.2203

Run it

Terminal

$ human-physiology solve hp_bmr_harris_benedict "weight=70 kg" "height=175 cm" age=30 sex=male version=original
BMR = 1702.0125
BMR_kJ = 7121.2203
  by: Basal metabolic rate by the Harris-Benedict equation  (published linear coefficients)
  weight = 70 kg
  height = 1.75 m
  age = 30
  sex = male
  version = original
  assumes: adult; weight, height > 0

import human_physiology_ai999x as m

a = m.solve("hp_bmr_harris_benedict", weight="70 kg", height="175 cm", age="30", sex="male", version="original")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Basal metabolic rate by the Katch-McArdle equation2 fig.hp_bmr_katch_mcardle

Lean body mass W (1 – body fat) and BMR = 370 + 21.6 LBM in kcal/day (computed value).

BMR = 370 + 21.6\,LBM

You give

	Meaning	Unit	Example
\text{weight}	Weight (e.g. in kg)	kg	70 kg
\text{body\_fat}	Body fat fraction	—	0.2

You get

	Meaning	Unit	Example
\text{LBM}	Lean body mass in kg	—	56
\text{BMR}	Basal metabolic rate in kcal/day	—	1579.6

Values with a unit may be written in any unit of the right kind.

Applies when

weight > 0
0 ≤ body fat < 1

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Katch-McArdle BMR: weight = 70 kg, body_fat = 0.2

LBM = 56

BMR = 1579.6

Run it

Terminal

$ human-physiology solve hp_bmr_katch_mcardle "weight=70 kg" body_fat=0.2
LBM = 56
BMR = 1579.6
  by: Basal metabolic rate by the Katch-McArdle equation  (Katch-McArdle)
  weight = 70 kg
  body_fat = 0.2
  assumes: weight > 0; 0 <= body fat < 1

import human_physiology_ai999x as m

a = m.solve("hp_bmr_katch_mcardle", weight="70 kg", body_fat="0.2")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Energy expenditure of an activity from METs1 fig.hp_met_energy

1 MET = 3.5 mL O2/kg/min; oxygen uptake MET x 3.5 x W and energy MET x 3.5 x W/200 kcal/min.

\text{kcal/min} = \frac{MET \times 3.5 \times W}{200}

You give

	Meaning	Unit	Example
\text{MET}	Activity intensity in METs	—	8
\text{weight}	Weight (e.g. in kg)	kg	70 kg
\text{duration}	Duration (e.g. in min)	s	30 min

You get

	Meaning	Unit	Example
\text{VO2}	Oxygen uptake in mL/min	—	1960
\text{kcal\_per\_min}	Energy in kcal/min	—	9.8
\text{kcal}	Total energy in kcal	—	294

Values with a unit may be written in any unit of the right kind.

Applies when

all inputs > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

MET energy expenditure: MET = 8, weight = 70 kg, duration = 30 min

VO2 = 1960

kcal_per_min = 9.8

kcal = 294

Run it

Terminal

$ human-physiology solve hp_met_energy MET=8 "weight=70 kg" "duration=30 min"
VO2 = 1960
kcal_per_min = 9.8
kcal = 294
  by: Energy expenditure of an activity from METs  (kcal/min = MET x 3.5 x W / 200 (about 5 kcal per L O2))
  MET = 8
  weight = 70 kg
  duration = 1800 s
  assumes: all inputs > 0

import human_physiology_ai999x as m

a = m.solve("hp_met_energy", MET="8", weight="70 kg", duration="30 min")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

First-order elimination: rate constant and half-life from clearance and volume1 fig.hp_elimination_half_life

Physiology of a one-compartment first-order process: k = CL/Vd, half-life ln 2/k and the fraction remaining after t (not a dosing recommendation).

t_{1/2} = \frac{\ln 2\, V_d}{CL}

You give

	Meaning	Unit	Example
\text{CL}	Clearance (e.g. in L/h)	m^3/s	5 L/h
\text{Vd}	Volume of distribution (e.g. in L)	m^3	50 L
t	Elapsed time (e.g. in h) (optional) optional	s	12 h

You get

	Meaning	Unit	Example
k	Elimination rate constant per hour	—	0.1
\text{half\_life}	Half-life in h	—	6.931471806
\text{fraction\_remaining}	Fraction remaining after t	—	0.3011942119

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

CL, Vd > 0
t >= 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

elimination half-life: CL = 5 L/h, Vd = 50 L, t = 12 h

k = 0.1

half_life = 6.931471806

fraction_remaining = 0.3011942119

Run it

Terminal

$ human-physiology solve hp_elimination_half_life "CL=5 L/h" "Vd=50 L" "t=12 h"
k = 0.1
half_life = 6.931471806
fraction_remaining = 0.3011942119
  by: First-order elimination: rate constant and half-life from clearance and volume  (k = CL/Vd; t1/2 = ln 2 / k; C/C0 = exp(-k t))
  CL = 1.388888889e-06 m^3/s
  Vd = 0.05 m^3
  t = 43200 s
  assumes: CL, Vd > 0; t >= 0

import human_physiology_ai999x as m

a = m.solve("hp_elimination_half_life", CL="5 L/h", Vd="50 L", t="12 h")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Food energy by the Atwater factors1 fig.hp_atwater_energy

Energy from macronutrients with the general Atwater factors 4 (carbohydrate), 9 (fat), 4 (protein) and 7 (alcohol) kcal/g.

E = 4C + 9F + 4P + 7A

You give

	Meaning	Unit	Example
\text{carbohydrate}	Carbohydrate (e.g. in g)	kg	300 g
\text{fat}	Fat (e.g. in g)	kg	70 g
\text{protein}	Protein (e.g. in g)	kg	80 g
\text{alcohol}	Alcohol (e.g. in g) (optional; none if omitted) optional	kg

You get

	Meaning	Unit	Example
\text{kcal}	Energy in kcal	—	2150
\text{kJ}	Energy in kJ	—	8995.6

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

amounts >= 0, not all zero

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Atwater energy: carbohydrate = 300 g, fat = 70 g, protein = 80 g

kcal = 2150

kJ = 8995.6

Run it

Terminal

$ human-physiology solve hp_atwater_energy "carbohydrate=300 g" "fat=70 g" "protein=80 g"
kcal = 2150
kJ = 8995.6
  by: Food energy by the Atwater factors  (4/9/4/7 kcal per g; 1 kcal = 4.184 kJ)
  carbohydrate = 0.3 kg
  fat = 0.07 kg
  protein = 0.08 kg
  assumes: amounts >= 0, not all zero

import human_physiology_ai999x as m

a = m.solve("hp_atwater_energy", carbohydrate="300 g", fat="70 g", protein="80 g")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Maximal heart rate, heart rate reserve and Karvonen target1 fig.hp_heart_rate_reserve

Estimated maximal heart rate (220 – age; Tanaka 208 – 0.7 age), the reserve HRmax – HRrest and the Karvonen target HRrest + intensity x reserve.

HR_{target} = HR_{rest} + I\,(HR_{max} – HR_{rest})

You give

	Meaning	Unit	Example
\text{age}	Age in years	—	40
\text{HR\_rest}	Resting heart rate in beats/min	—	60
\text{intensity}	Exercise intensity (fraction of reserve)	—	0.7

You get

	Meaning	Unit	Example
\text{HRmax}	Age-predicted maximal heart rate (Fox formula) in beats/min	—	180
\text{HRmax\_Tanaka}	Age-predicted maximal heart rate (Tanaka formula) in beats/min	—	180
\text{HRR}	Heart rate reserve in beats/min	—	120
\text{target}	Karvonen target heart rate in beats/min	—	144

Applies when

5 ≤ age <= 100
HR_rest below HRmax

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Karvonen target heart rate: age = 40, HR_rest = 60, intensity = 0.7

HRmax = 180

HRmax_Tanaka = 180

HRR = 120

target = 144

Run it

Terminal

$ human-physiology solve hp_heart_rate_reserve age=40 HR_rest=60 intensity=0.7
HRmax = 180
HRmax_Tanaka = 180
HRR = 120
target = 144
  by: Maximal heart rate, heart rate reserve and Karvonen target  (Karvonen with HRmax = 220 - age)
  age = 40
  HR_rest = 60
  intensity = 0.7
  assumes: 5 <= age <= 100; HR_rest below HRmax

import human_physiology_ai999x as m

a = m.solve("hp_heart_rate_reserve", age="40", HR_rest="60", intensity="0.7")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Respiratory quotient and energy expenditure (Weir equation)2 fig.hp_weir_energy_expenditure

RQ = VCO2/VO2 and resting energy expenditure by the abbreviated Weir equation (3.941 VO2 + 1.106 VCO2) x 1440 kcal/day (L/min).

EE = (3.941\,\dot V_{O_2} + 1.106\,\dot V_{CO_2})\times 1440

You give

	Meaning	Unit	Example
\text{VO2}	Oxygen consumption (e.g. in mL/min)	m^3/s	250 mL/min
\text{VCO2}	CO2 production (e.g. in mL/min)	m^3/s	200 mL/min

You get

	Meaning	Unit	Example
\text{RQ}	Respiratory quotient	—	0.8
\text{EE}	Energy expenditure in kcal/day	—	1737.288
\text{EE\_kJ}	Energy expenditure in kJ/day	—	7268.812992

Values with a unit may be written in any unit of the right kind.

Applies when

VO2, VCO2 > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Weir energy expenditure: VO2 = 250 mL/min, VCO2 = 200 mL/min

RQ = 0.8

EE = 1737.288

EE_kJ = 7268.812992

Run it

Terminal

$ human-physiology solve hp_weir_energy_expenditure "VO2=250 mL/min" "VCO2=200 mL/min"
RQ = 0.8
EE = 1737.288
EE_kJ = 7268.812992
  by: Respiratory quotient and energy expenditure (Weir equation)  (Weir (1949) without urinary nitrogen)
  VO2 = 4.166666667e-06 m^3/s
  VCO2 = 3.333333333e-06 m^3/s
  assumes: VO2, VCO2 > 0

import human_physiology_ai999x as m

a = m.solve("hp_weir_energy_expenditure", VO2="250 mL/min", VCO2="200 mL/min")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Resting energy expenditure by the Mifflin-St Jeor equation1 fig.hp_bmr_mifflin_st_jeor

Mifflin-St Jeor (1990): 10 W + 6.25 H – 5 A + 5 (male) or – 161 (female), kcal/day (computed value).

REE = 10W + 6.25H – 5A + s

You give

	Meaning	Unit	Example
\text{weight}	Weight (e.g. in kg)	kg	70 kg
\text{height}	Height (e.g. in cm)	m	175 cm
\text{age}	Age in years	—	30
\text{sex}	Sex for the constant choice	—	male

You get

	Meaning	Unit	Example
\text{REE}	Resting energy expenditure in kcal/day	—	1648.75
\text{REE\_kJ}	Resting energy expenditure in kJ/day	—	6898.37

Values with a unit may be written in any unit of the right kind.

Applies when

adult
weight, height > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Mifflin-St Jeor: weight = 70 kg, height = 175 cm, age = 30, sex = male

REE = 1648.75

REE_kJ = 6898.37

Run it

Terminal

$ human-physiology solve hp_bmr_mifflin_st_jeor "weight=70 kg" "height=175 cm" age=30 sex=male
REE = 1648.75
REE_kJ = 6898.37
  by: Resting energy expenditure by the Mifflin-St Jeor equation  (Mifflin-St Jeor)
  weight = 70 kg
  height = 1.75 m
  age = 30
  sex = male
  assumes: adult; weight, height > 0

import human_physiology_ai999x as m

a = m.solve("hp_bmr_mifflin_st_jeor", weight="70 kg", height="175 cm", age="30", sex="male")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Nerve and muscle 12

Charge and number of ions to change the membrane potential1 fig.hp_membrane_charge

Q = Cm A dV and the number of monovalent ions it represents (why few ions cross in an action potential).

Q = C_m A \Delta V,\quad N = \frac{|Q|}{e}

You give

	Meaning	Unit	Example
\text{Cm}	Specific membrane capacitance (e.g. in uF/cm^2)	F/m^2	1 uF/cm^2
\text{dV}	Voltage change (e.g. in mV)	V	100 mV
\text{area}	Membrane area (e.g. in um²)	m^2	1000 um^2

You get

	Meaning	Unit	Example
Q_{\mathrm{pC}}	Charge in pC	—	1
\text{ions}	Number of monovalent ions	—	6241509.074

Values with a unit may be written in any unit of the right kind.

Applies when

Cm, area > 0
dV not 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

membrane charge: Cm = 1 uF/cm^2, dV = 100 mV, area = 1000 um^2

Q_pC = 1

ions = 6241509.074

Run it

Terminal

$ human-physiology solve hp_membrane_charge "Cm=1 uF/cm^2" "dV=100 mV" "area=1000 um^2"
Q_pC = 1
ions = 6241509.074
  by: Charge and number of ions to change the membrane potential  (Q = C A dV; ions = |Q|/e)
  Cm = 0.01 F/m^2
  dV = 0.1 V
  area = 1e-09 m^2
  assumes: Cm, area > 0; dV not 0

import human_physiology_ai999x as m

a = m.solve("hp_membrane_charge", Cm="1 uF/cm^2", dV="100 mV", area="1000 um^2")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Conduction velocity scaling with fibre diameter1 fig.hp_conduction_velocity_scaling

Conduction velocity is proportional to diameter for myelinated fibres and to its square root for unmyelinated fibres.

v \propto d \;(\text{myelinated}),\quad v \propto \sqrt{d}\;(\text{unmyelinated})

You give

	Meaning	Unit	Example
\text{v1}	Known conduction velocity (e.g. in m/s)	m/s	60 m/s
\text{d1}	Diameter of that fibre (e.g. in um)	m	10 um
\text{d2}	Diameter of the other fibre (e.g. in um)	m	5 um
\text{fiber}	Fibre type choice	—	myelinated

You get

	Meaning	Unit	Example
\text{v2}	Conduction velocity of the other fibre in m/s	—	30
\text{ratio}	Ratio of the new to the known velocity	—	0.5

Values with a unit may be written in any unit of the right kind.

Applies when

all inputs > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

conduction velocity scaling: v1 = 60 m/s, d1 = 10 um, d2 = 5 um, fiber = myelinated

v2 = 30

ratio = 0.5

Run it

Terminal

$ human-physiology solve hp_conduction_velocity_scaling "v1=60 m/s" "d1=10 um" "d2=5 um" fiber=myelinated
v2 = 30
ratio = 0.5
  by: Conduction velocity scaling with fibre diameter  (v ~ d (myelinated), v ~ sqrt(d) (unmyelinated, cable theory))
  v1 = 60 m/s
  d1 = 1e-05 m
  d2 = 5e-06 m
  fiber = myelinated
  assumes: all inputs > 0

import human_physiology_ai999x as m

a = m.solve("hp_conduction_velocity_scaling", v1="60 m/s", d1="10 um", d2="5 um", fiber="myelinated")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Hill muscle force-velocity relation1 fig.hp_hill_force_velocity

Hill's equation (F + a)(v + b) = (F0 + a) b: force at shortening velocity v, the power F v and vmax = b F0/a.

(F + a)(v + b) = (F_0 + a)\,b

You give

	Meaning	Unit	Example
\text{F0}	Isometric force (e.g. in N)	N	100 N
a	Hill constant a (e.g. in N)	N	25 N
b	Hill constant b (e.g. in m/s)	m/s	0.3 m/s
v	Shortening velocity (e.g. in m/s)	m/s	0.1 m/s

You get

	Meaning	Unit	Example
F	Force in N	—	68.75
\text{power}	Power in W	—	6.875
\text{vmax}	Maximal shortening velocity in m/s	—	1.2

Values with a unit may be written in any unit of the right kind.

Applies when

F0, a, b > 0
0 <= v <= vmax

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Hill force-velocity: F0 = 100 N, a = 25 N, b = 0.3 m/s, v = 0.1 m/s

F = 68.75

power = 6.875

vmax = 1.2

Run it

Terminal

$ human-physiology solve hp_hill_force_velocity "F0=100 N" "a=25 N" "b=0.3 m/s" "v=0.1 m/s"
F = 68.75
power = 6.875
vmax = 1.2
  by: Hill muscle force-velocity relation  (F = (F0 b - a v)/(v + b))
  F0 = 100 N
  a = 25 N
  b = 0.3 m/s
  v = 0.1 m/s
  assumes: F0, a, b > 0; 0 <= v <= vmax

import human_physiology_ai999x as m

a = m.solve("hp_hill_force_velocity", F0="100 N", a="25 N", b="0.3 m/s", v="0.1 m/s")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Ionic current and driving force1 fig.hp_ionic_current

Ohmic ionic current I = g (Vm – E) (positive = outward) and the driving force Vm – E.

I = g\,(V_m – E)

You give

	Meaning	Unit	Example
g	Conductance (e.g. in nS)	S	10 nS
\text{Vm}	Membrane potential (e.g. in mV)	V	-70 mV
E	Equilibrium potential of the ion (e.g. in mV)	V	-95 mV

You get

	Meaning	Unit	Example
I	Current in pA (positive outward)	—	250
\text{driving\_force}	Driving force in mV	—	25

Values with a unit may be written in any unit of the right kind.

Applies when

g >= 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

ionic current: g = 10 nS, Vm = -70 mV, E = -95 mV

I = 250

driving_force = 25

Run it

Terminal

$ human-physiology solve hp_ionic_current "g=10 nS" "Vm=-70 mV" "E=-95 mV"
I = 250
driving_force = 25
  by: Ionic current and driving force  (I = g (Vm - E); nS x mV = pA)
  g = 1e-08 S
  Vm = -0.07 V
  E = -0.095 V
  assumes: g >= 0

import human_physiology_ai999x as m

a = m.solve("hp_ionic_current", g="10 nS", Vm="-70 mV", E="-95 mV")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Length constant of an axon and steady-state attenuation1 fig.hp_length_constant

lambda = sqrt(Rm d/(4 Ri)) of a passive fibre and the steady voltage fraction e^(-x/lambda) at distance x.

\lambda = \sqrt{\frac{R_m d}{4 R_i}},\quad V(x) = V_0 e^{-x/\lambda}

You give

	Meaning	Unit	Example
\text{Rm}	Specific membrane resistance in ohm cm²	—	20000
\text{Ri}	Axoplasm resistivity in ohm cm	—	100
d	Fibre diameter (e.g. in um)	m	10 um
x	Distance from the injection site (e.g. in mm) (optional; needed for the attenuation) optional	m	1 mm

You get

	Meaning	Unit	Example
\lambda	Length constant in mm	—	2.236067977
\text{attenuation}	Steady-state fraction of the voltage remaining at distance x	—	0.6394073192

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

Rm, Ri, d > 0
x >= 0
semi-infinite passive cable

Outside these conditions it says what is wrong instead of returning a number.

Worked example

length constant: Rm = 20000, Ri = 100, d = 10 um, x = 1 mm

lambda = 2.236067977

attenuation = 0.6394073192

Run it

Terminal

$ human-physiology solve hp_length_constant Rm=20000 Ri=100 "d=10 um" "x=1 mm"
lambda = 2.236067977
attenuation = 0.6394073192
  by: Length constant of an axon and steady-state attenuation  (lambda = sqrt(rm/ri))
  Rm = 20000
  Ri = 100
  d = 1e-05 m
  x = 0.001 m
  assumes: Rm, Ri, d > 0; x >= 0; semi-infinite passive cable

import human_physiology_ai999x as m

a = m.solve("hp_length_constant", Rm="20000", Ri="100", d="10 um", x="1 mm")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Maximal power of a muscle from Hill's equation1 fig.hp_muscle_max_power

The shortening velocity of maximal power v_opt = b (sqrt(1 + F0/a) – 1), the force there and the maximal power.

v_{opt} = b\left(\sqrt{1 + F_0/a} – 1\right)

You give

	Meaning	Unit	Example
\text{F0}	Isometric force (e.g. in N)	N	100 N
a	Hill constant a (e.g. in N)	N	25 N
b	Hill constant b (e.g. in m/s)	m/s	0.3 m/s

You get

	Meaning	Unit	Example
v_{\mathrm{opt}}	Velocity of maximal power in m/s	—	0.3708203932
F_{\mathrm{opt}}	Force at maximal power in N	—	30.90169944
P_{\mathrm{max}}	Maximal power in W	—	11.45898034
\text{vmax}	Maximal shortening velocity in m/s	—	1.2

Values with a unit may be written in any unit of the right kind.

Applies when

F0, a, b > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

muscle maximal power: F0 = 100 N, a = 25 N, b = 0.3 m/s

v_opt = 0.3708203932

F_opt = 30.90169944

P_max = 11.45898034

vmax = 1.2

Run it

Terminal

$ human-physiology solve hp_muscle_max_power "F0=100 N" "a=25 N" "b=0.3 m/s"
v_opt = 0.3708203932
F_opt = 30.90169944
P_max = 11.45898034
vmax = 1.2
  by: Maximal power of a muscle from Hill's equation  (dP/dv = 0 for P = v (F0 b - a v)/(v + b))
  F0 = 100 N
  a = 25 N
  b = 0.3 m/s
  assumes: F0, a, b > 0

import human_physiology_ai999x as m

a = m.solve("hp_muscle_max_power", F0="100 N", a="25 N", b="0.3 m/s")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Membrane potential by the chord conductance equation1 fig.hp_chord_conductance

Steady potential at which the ohmic ionic currents cancel: Vm = (gK EK + gNa ENa + gCl ECl)/(gK + gNa + gCl).

V_m = \frac{g_K E_K + g_{Na}E_{Na} + g_{Cl}E_{Cl}}{g_K + g_{Na} + g_{Cl}}

You give

	Meaning	Unit	Example
\text{gK}	K+ conductance (any unit)	—	1
\text{gNa}	Na+ conductance (same unit)	—	0.05
\text{gCl}	Cl- conductance (same unit)	—	0.3
\text{EK}	K+ equilibrium potential (e.g. in mV)	V	-95 mV
\text{ENa}	Na+ equilibrium potential (e.g. in mV)	V	67 mV
\text{ECl}	Cl- equilibrium potential (e.g. in mV)	V	-90 mV

You get

	Meaning	Unit	Example
\text{Vm}	Membrane potential in mV	—	-87.88888889

Values with a unit may be written in any unit of the right kind.

Applies when

conductances >= 0, not all zero

Outside these conditions it says what is wrong instead of returning a number.

Worked example

chord conductance: gK = 1, gNa = 0.05, gCl = 0.3, EK = -95 mV, ENa = 67 mV, ECl = -90 mV

Vm = -87.88888889

Run it

Terminal

$ human-physiology solve hp_chord_conductance gK=1 gNa=0.05 gCl=0.3 "EK=-95 mV" "ENa=67 mV" "ECl=-90 mV"
Vm = -87.88888889
  by: Membrane potential by the chord conductance equation  (conductance-weighted mean of the equilibrium potentials)
  gK = 1
  gNa = 0.05
  gCl = 0.3
  EK = -0.095 V
  ENa = 0.067 V
  ECl = -0.09 V
  assumes: conductances >= 0, not all zero

import human_physiology_ai999x as m

a = m.solve("hp_chord_conductance", gK="1", gNa="0.05", gCl="0.3", EK="-95 mV", ENa="67 mV", ECl="-90 mV")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Membrane time constant and charging1 fig.hp_membrane_time_constant

tau_m = Rm Cm (specific membrane resistance times capacitance) and the fraction of the final voltage reached after t.

\tau_m = R_m C_m

You give

	Meaning	Unit	Example
\text{Rm}	Specific membrane resistance in ohm cm²	—	20000
\text{Cm}	Specific membrane capacitance (e.g. in uF/cm^2)	F/m^2	1 uF/cm^2
t	Time (e.g. in ms) (optional; needed for the fraction charged) optional	s	10 ms

You get

	Meaning	Unit	Example
\tau	Membrane time constant in ms	—	20
\text{fraction}	Fraction of the final voltage at time t	—	0.3934693403

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

Rm, Cm > 0
t >= 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

membrane time constant: Rm = 20000, Cm = 1 uF/cm^2, t = 10 ms

tau = 20

fraction = 0.3934693403

Run it

Terminal

$ human-physiology solve hp_membrane_time_constant Rm=20000 "Cm=1 uF/cm^2" "t=10 ms"
tau = 20
fraction = 0.3934693403
  by: Membrane time constant and charging  (tau = Rm Cm; V(t)/V_inf = 1 - exp(-t/tau))
  Rm = 20000
  Cm = 0.01 F/m^2
  t = 0.01 s
  assumes: Rm, Cm > 0; t >= 0

import human_physiology_ai999x as m

a = m.solve("hp_membrane_time_constant", Rm="20000", Cm="1 uF/cm^2", t="10 ms")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Nernst equilibrium potential of an ion2 fig.hp_nernst_potential

E = (RT/zF) ln(Cout/Cin) in mV for an ion of valence z.

E = \frac{RT}{zF}\ln\frac{[C]_o}{[C]_i}

You give

	Meaning	Unit	Example
z	Ion valence (charge number, negative for anions)	—	1
\text{Cout}	Extracellular concentration in mM	—	4
\text{Cin}	Intracellular concentration in mM	—	140
T	Temperature (e.g. in degC) (optional; body temperature if omitted) optional	K

You get

	Meaning	Unit	Example
E	Equilibrium potential in mV	—	-95.02257566
\text{RT\_F}	RT/F in mV	—	26.72665911

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

z a nonzero whole number
concentrations > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Nernst potential: z = 1, Cout = 4, Cin = 140

E = -95.02257566

RT_F = 26.72665911

Run it

Terminal

$ human-physiology solve hp_nernst_potential z=1 Cout=4 Cin=140
E = -95.02257566
RT_F = 26.72665911
  by: Nernst equilibrium potential of an ion  (Nernst equation)
  Cout = 4
  Cin = 140
  assumes: z a nonzero whole number; concentrations > 0

import human_physiology_ai999x as m

a = m.solve("hp_nernst_potential", z="1", Cout="4", Cin="140")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Nerve conduction velocity from two stimulation sites1 fig.hp_nerve_conduction_velocity

NCV = distance between stimulation sites / difference of the response latencies (mm/ms = m/s).

NCV = \frac{d}{t_p – t_d}

You give

	Meaning	Unit	Example
\text{distance}	Distance between stimulation sites (e.g. in mm)	m	250 mm
\text{latency\_proximal}	Latency from the proximal site (e.g. in ms)	s	8.5 ms
\text{latency\_distal}	Latency from the distal site (e.g. in ms)	s	3.5 ms

You get

	Meaning	Unit	Example
\text{NCV}	Conduction velocity in m/s	—	50

Values with a unit may be written in any unit of the right kind.

Applies when

latency_proximal > latency_distal

Outside these conditions it says what is wrong instead of returning a number.

Worked example

nerve conduction velocity: distance = 250 mm, latency_proximal = 8.5 ms, latency_distal = 3.5 ms

NCV = 50

Run it

Terminal

$ human-physiology solve hp_nerve_conduction_velocity "distance=250 mm" "latency_proximal=8.5 ms" "latency_distal=3.5 ms"
NCV = 50
  by: Nerve conduction velocity from two stimulation sites  (NCV = d / (t_p - t_d))
  distance = 0.25 m
  latency_proximal = 0.0085 s
  latency_distal = 0.0035 s
  assumes: latency_proximal > latency_distal

import human_physiology_ai999x as m

a = m.solve("hp_nerve_conduction_velocity", distance="250 mm", latency_proximal="8.5 ms", latency_distal="3.5 ms")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Resting membrane potential by the Goldman-Hodgkin-Katz equation1 fig.hp_ghk_voltage

Vm = (RT/F) ln((PK K_o + PNa Na_o + PCl Cl_i)/(PK K_i + PNa Na_i + PCl Cl_o)) with relative permeabilities.

V_m = \frac{RT}{F}\ln\frac{P_K[K^+]_o + P_{Na}[Na^+]_o + P_{Cl}[Cl^-]_i}{P_K[K^+]_i + P_{Na}[Na^+]_i + P_{Cl}[Cl^-]_o}

You give

	Meaning	Unit	Example
\text{PK}	Relative K+ permeability	—	1
\text{PNa}	Relative Na+ permeability	—	0.04
\text{PCl}	Relative Cl- permeability	—	0.45
K_{o}	Extracellular K+ in mM	—	4
K_{i}	Intracellular K+ in mM	—	140
\text{Na\_o}	Extracellular Na+ in mM	—	145
\text{Na\_i}	Intracellular Na+ in mM	—	12
\text{Cl\_o}	Extracellular Cl- in mM	—	116
\text{Cl\_i}	Intracellular Cl- in mM	—	4.2
T	Temperature (e.g. in degC) (optional; body temperature if omitted) optional	K

You get

	Meaning	Unit	Example
\text{Vm}	Membrane potential in mV	—	-74.89603682

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

PK > 0
PNa, PCl >= 0
concentrations > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

GHK voltage: PK = 1, PNa = 0.04, PCl = 0.45, K_o = 4, K_i = 140, Na_o = 145, Na_i = 12, Cl_o = 116, Cl_i = 4.2

Vm = -74.89603682

Run it

Terminal

$ human-physiology solve hp_ghk_voltage PK=1 PNa=0.04 PCl=0.45 K_o=4 K_i=140 Na_o=145 Na_i=12 Cl_o=116 Cl_i=4.2
Vm = -74.89603682
  by: Resting membrane potential by the Goldman-Hodgkin-Katz equation  (GHK voltage equation (chloride terms inverted))
  PK = 1
  PNa = 0.04
  PCl = 0.45
  K_o = 4
  K_i = 140
  Na_o = 145
  Na_i = 12
  Cl_o = 116
  Cl_i = 4.2
  assumes: PK > 0; PNa, PCl >= 0; concentrations > 0

import human_physiology_ai999x as m

a = m.solve("hp_ghk_voltage", PK="1", PNa="0.04", PCl="0.45", K_o="4", K_i="140", Na_o="145", Na_i="12", Cl_o="116", Cl_i="4.2")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Threshold current for a pulse width (strength-duration curve)1 fig.hp_strength_duration

Threshold current for a pulse of width t from rheobase and chronaxie: Lapicque (exponential RC) and Weiss (linear charge) forms.

I = \frac{I_{rh}}{1 – 2^{-t/c}}\;(\text{Lapicque}),\quad I = I_{rh}\left(1 + \frac{c}{t}\right)\;(\text{Weiss})

You give

	Meaning	Unit	Example
\text{rheobase}	Rheobase current (any unit, e.g. mA)	—	1
\text{chronaxie}	Chronaxie (e.g. in ms)	s	0.5 ms
t	Pulse width (e.g. in ms)	s	0.5 ms

You get

	Meaning	Unit	Example
I_{\mathrm{lapicque}}	Threshold current by the Lapicque (exponential) law, in the unit of the rheobase	—	2
I_{\mathrm{weiss}}	Threshold current by the Weiss (linear) law, in the unit of the rheobase	—	2

Values with a unit may be written in any unit of the right kind.

Applies when

all inputs > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

strength-duration threshold: rheobase = 1, chronaxie = 0.5 ms, t = 0.5 ms

I_lapicque = 2

I_weiss = 2

Run it

Terminal

$ human-physiology solve hp_strength_duration rheobase=1 "chronaxie=0.5 ms" "t=0.5 ms"
I_lapicque = 2
I_weiss = 2
  by: Threshold current for a pulse width (strength-duration curve)  (Lapicque: tau = chronaxie/ln 2; Weiss: linear charge law)
  rheobase = 1
  chronaxie = 0.0005 s
  t = 0.0005 s
  assumes: all inputs > 0

import human_physiology_ai999x as m

a = m.solve("hp_strength_duration", rheobase="1", chronaxie="0.5 ms", t="0.5 ms")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Renal 12

Calculated and effective plasma osmolality and the osmolal gap2 fig.hp_plasma_osmolality

Calculated osmolality 2 Na + glucose/18 + BUN/2.8, effective osmolality (tonicity) 2 Na + glucose/18 and, with a measured value, the osmolal gap.

P_{osm} = 2[Na^+] + \frac{glucose}{18} + \frac{BUN}{2.8}

You give

	Meaning	Unit	Example
\text{Na}	Sodium (e.g. in mmol/L)	mol/m^3	140 mmol/L
\text{glucose}	Glucose in mg/dL	—	90
\text{BUN}	Blood urea nitrogen in mg/dL	—	14
\text{measured}	Measured osmolality in mOsm/kg (optional) optional	—

You get

	Meaning	Unit	Example
\text{calculated}	Calculated osmolality in mOsm/kg	—	290
\text{effective}	Effective osmolality in mOsm/kg	—	285
\text{osmolal\_gap}	Measured – calculated in mOsm/kg	—

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

Na > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

plasma osmolality: Na = 140 mmol/L, glucose = 90, BUN = 14

calculated = 290

effective = 285

Run it

Terminal

$ human-physiology solve hp_plasma_osmolality "Na=140 mmol/L" glucose=90 BUN=14
calculated = 290
effective = 285
  by: Calculated and effective plasma osmolality and the osmolal gap  (published divisors 18 (glucose) and 2.8 (urea nitrogen))
  Na = 140 mol/m^3
  glucose = 90
  BUN = 14
  assumes: Na > 0

import human_physiology_ai999x as m

a = m.solve("hp_plasma_osmolality", Na="140 mmol/L", glucose="90", BUN="14")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Creatinine clearance by the Cockcroft-Gault equation1 fig.hp_cockcroft_gault

Estimated creatinine clearance (140 – age) x weight/(72 x Scr), x 0.85 for female (computed value only).

C_{Cr} = \frac{(140 – age)\,W}{72\,S_{cr}}\;[\times 0.85]

You give

	Meaning	Unit	Example
\text{age}	Age in years	—	60
\text{weight}	Body weight (e.g. in kg)	kg	72 kg
\text{Scr}	Serum creatinine in mg/dL	—	1
\text{sex}	Sex for the equation's coefficient choice	—	male

You get

	Meaning	Unit	Example
\text{CrCl}	Estimated creatinine clearance in mL/min	—	80

Values with a unit may be written in any unit of the right kind.

Applies when

18 ≤ age < 140
weight, Scr > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Cockcroft-Gault creatinine clearance: age = 60, weight = 72 kg, Scr = 1, sex = male

CrCl = 80

Run it

Terminal

$ human-physiology solve hp_cockcroft_gault age=60 "weight=72 kg" Scr=1 sex=male
CrCl = 80
  by: Creatinine clearance by the Cockcroft-Gault equation  (Cockcroft-Gault (1976))
  age = 60
  weight = 72 kg
  Scr = 1
  sex = male
  assumes: 18 <= age < 140; weight, Scr > 0

import human_physiology_ai999x as m

a = m.solve("hp_cockcroft_gault", age="60", weight="72 kg", Scr="1", sex="male")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Creatinine clearance from a timed urine collection1 fig.hp_creatinine_clearance_timed

Urine flow from a timed collection, creatinine clearance U V/P and its value normalized to 1.73 m^2.

C_{Cr} = \frac{U_{Cr}\dot V}{P_{Cr}}

You give

	Meaning	Unit	Example
\text{Ucr}	Urine creatinine in mg/dL	—	100
\text{Vurine}	Urine volume collected (e.g. in mL)	m^3	1440 mL
\text{hours}	Collection time (e.g. in h)	s	24 h
\text{Pcr}	Plasma creatinine in mg/dL	—	1
\text{BSA}	Body surface area (e.g. in m²) (optional; standard adult area if omitted) optional	m^2

You get

	Meaning	Unit	Example
\text{urine\_flow}	Urine flow in mL/min	—	1
\text{CrCl}	Creatinine clearance in mL/min	—	100
\text{CrCl\_normalized}	Creatinine clearance in mL/min normalized to standard body surface area	—	100

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

all inputs > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

creatinine clearance from a timed urine: Ucr = 100, Vurine = 1440 mL, hours = 24 h, Pcr = 1

urine_flow = 1

CrCl = 100

CrCl_normalized = 100

Run it

Terminal

$ human-physiology solve hp_creatinine_clearance_timed Ucr=100 "Vurine=1440 mL" "hours=24 h" Pcr=1
urine_flow = 1
CrCl = 100
CrCl_normalized = 100
  by: Creatinine clearance from a timed urine collection  (V = volume/(60 hours); CrCl = U V/P; x 1.73/BSA)
  Ucr = 100
  Vurine = 0.00144 m^3
  hours = 86400 s
  Pcr = 1
  assumes: all inputs > 0

import human_physiology_ai999x as m

a = m.solve("hp_creatinine_clearance_timed", Ucr="100", Vurine="1440 mL", hours="24 h", Pcr="1")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Estimated GFR by the CKD-EPI 2021 creatinine equation2 fig.hp_egfr_ckd_epi_2021

eGFR = 142 min(Scr/k, 1)^a max(Scr/k, 1)^-1.200 0.9938^age (x 1.012 female), the race-free 2021 CKD-EPI creatinine equation (computed value only).

eGFR = 142\,\min(S_{cr}/\kappa,1)^{\alpha}\max(S_{cr}/\kappa,1)^{-1.200}\,0.9938^{age}\,[1.012]

You give

	Meaning	Unit	Example
\text{Scr}	Serum creatinine in mg/dL	—	1
\text{age}	Age in years	—	50
\text{sex}	Sex for the equation's coefficients choice	—	male

You get

	Meaning	Unit	Example
\text{eGFR}	Estimated GFR in mL/min normalized to standard body surface area	—	91.69147861

Applies when

Scr > 0
18 ≤ age <= 120

Outside these conditions it says what is wrong instead of returning a number.

Worked example

eGFR CKD-EPI 2021: Scr = 1, age = 50, sex = male

eGFR = 91.69147861

Run it

Terminal

$ human-physiology solve hp_egfr_ckd_epi_2021 Scr=1 age=50 sex=male
eGFR = 91.69147861
  by: Estimated GFR by the CKD-EPI 2021 creatinine equation  (CKD-EPI 2021: k = 0.7/0.9, a = -0.241/-0.302 (female/male))
  Scr = 1
  age = 50
  sex = male
  assumes: Scr > 0; 18 <= age <= 120

import human_physiology_ai999x as m

a = m.solve("hp_egfr_ckd_epi_2021", Scr="1", age="50", sex="male")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Estimated GFR by the CKD-EPI 2021 creatinine-cystatin C equation1 fig.hp_egfr_ckd_epi_2021_cys

The race-free 2021 CKD-EPI equation from serum creatinine and cystatin C (computed value only).

eGFR = 135\,\min(S_{cr}/\kappa,1)^{\alpha}\max(S_{cr}/\kappa,1)^{-0.544}\min(S_{cys}/0.8,1)^{-0.323}\max(S_{cys}/0.8,1)^{-0.778}\,0.9961^{age}

You give

	Meaning	Unit	Example
\text{Scr}	Serum creatinine in mg/dL	—	1
\text{Scys}	Serum cystatin C (e.g. in mg/L)	kg/m^3	1 mg/L
\text{age}	Age in years	—	50
\text{sex}	Sex for the equation's coefficients choice	—	male

You get

	Meaning	Unit	Example
\text{eGFR}	Estimated GFR in mL/min normalized to standard body surface area	—	88.14399054

Values with a unit may be written in any unit of the right kind.

Applies when

Scr, Scys > 0
18 ≤ age <= 120

Outside these conditions it says what is wrong instead of returning a number.

Worked example

eGFR creatinine-cystatin C: Scr = 1, Scys = 1 mg/L, age = 50, sex = male

eGFR = 88.14399054

Run it

Terminal

$ human-physiology solve hp_egfr_ckd_epi_2021_cys Scr=1 "Scys=1 mg/L" age=50 sex=male
eGFR = 88.14399054
  by: Estimated GFR by the CKD-EPI 2021 creatinine-cystatin C equation  (135 min(Scr/k,1)^a max(Scr/k,1)^-0.544 min(Scys/0.8,1)^-0.323 max(Scys/0.8,1)^-0.778 0.9961^age (x 0.963 female))
  Scr = 1
  Scys = 0.001 kg/m^3
  age = 50
  sex = male
  assumes: Scr, Scys > 0; 18 <= age <= 120

import human_physiology_ai999x as m

a = m.solve("hp_egfr_ckd_epi_2021_cys", Scr="1", Scys="1 mg/L", age="50", sex="male")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Filtered load, excretion and net tubular reabsorption or secretion1 fig.hp_tubular_handling

Filtered load GFR x Px, excretion rate Ux x V and net reabsorption (positive) or secretion (negative) as their difference.

R_x = GFR\cdot P_x – U_x \dot V

You give

	Meaning	Unit	Example
\text{GFR}	GFR (e.g. in mL/min)	m^3/s	120 mL/min
\text{Px}	Plasma concentration in mg/dL	—	100
\text{Ux}	Urine concentration in mg/dL	—	0
V	Urine flow (e.g. in mL/min)	m^3/s	1 mL/min

You get

	Meaning	Unit	Example
\text{filtered\_load}	Filtered load in mg/min	—	120
\text{excreted}	Excretion rate in mg/min	—	0
\text{reabsorbed}	Net reabsorption in mg/min (negative = net secretion)	—	120

Values with a unit may be written in any unit of the right kind.

Applies when

GFR, V > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

tubular reabsorption: GFR = 120 mL/min, Px = 100, Ux = 0, V = 1 mL/min

filtered_load = 120

excreted = 0

reabsorbed = 120

Run it

Terminal

$ human-physiology solve hp_tubular_handling "GFR=120 mL/min" Px=100 Ux=0 "V=1 mL/min"
filtered_load = 120
excreted = 0
reabsorbed = 120
  by: Filtered load, excretion and net tubular reabsorption or secretion  (filtered - excreted (mg/dL x mL/min / 100 = mg/min))
  GFR = 2e-06 m^3/s
  Px = 100
  Ux = 0
  V = 1.666666667e-08 m^3/s
  assumes: GFR, V > 0

import human_physiology_ai999x as m

a = m.solve("hp_tubular_handling", GFR="120 mL/min", Px="100", Ux="0", V="1 mL/min")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Filtration fraction2 fig.hp_filtration_fraction

The fraction of renal plasma flow filtered at the glomerulus, FF = GFR/RPF.

FF = \frac{GFR}{RPF}

You give

	Meaning	Unit	Example
\text{GFR}	Glomerular filtration rate (e.g. in mL/min)	m^3/s	120 mL/min
\text{RPF}	Renal plasma flow (e.g. in mL/min)	m^3/s	600 mL/min

You get

	Meaning	Unit	Example
\text{FF}	Filtration fraction	—	0.2
\text{FF\_percent}	Filtration fraction in %	—	20

Values with a unit may be written in any unit of the right kind.

Applies when

0 < GFR < RPF

Outside these conditions it says what is wrong instead of returning a number.

Worked example

filtration fraction: GFR = 120 mL/min, RPF = 600 mL/min

FF = 0.2

FF_percent = 20

Run it

Terminal

$ human-physiology solve hp_filtration_fraction "GFR=120 mL/min" "RPF=600 mL/min"
FF = 0.2
FF_percent = 20
  by: Filtration fraction  (FF = GFR / RPF)
  GFR = 2e-06 m^3/s
  RPF = 1e-05 m^3/s
  assumes: 0 < GFR < RPF

import human_physiology_ai999x as m

a = m.solve("hp_filtration_fraction", GFR="120 mL/min", RPF="600 mL/min")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Fractional excretion of a solute1 fig.hp_fractional_excretion

FE = (Ux Pcr)/(Px Ucr): the fraction of the filtered load excreted, with creatinine clearance as GFR (e.g. FENa).

FE_x = \frac{U_x P_{Cr}}{P_x U_{Cr}}

You give

	Meaning	Unit	Example
\text{Ux}	Urine concentration of the solute (any unit)	—	40
\text{Px}	Plasma concentration of the solute (same unit)	—	140
\text{Ucr}	Urine creatinine (any unit)	—	100
\text{Pcr}	Plasma creatinine (same unit)	—	1

You get

	Meaning	Unit	Example
\text{FE}	Fractional excretion	—	0.002857142857
\text{FE\_percent}	Fractional excretion in %	—	0.2857142857

Applies when

Px, Ucr, Pcr > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

fractional excretion: Ux = 40, Px = 140, Ucr = 100, Pcr = 1

FE = 0.002857142857

FE_percent = 0.2857142857

Run it

Terminal

$ human-physiology solve hp_fractional_excretion Ux=40 Px=140 Ucr=100 Pcr=1
FE = 0.002857142857
FE_percent = 0.2857142857
  by: Fractional excretion of a solute  (FE = Ux Pcr / (Px Ucr))
  Ux = 40
  Px = 140
  Ucr = 100
  Pcr = 1
  assumes: Px, Ucr, Pcr > 0

import human_physiology_ai999x as m

a = m.solve("hp_fractional_excretion", Ux="40", Px="140", Ucr="100", Pcr="1")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Glucose titration curve (ideal transport maximum)1 fig.hp_glucose_titration

Filtered, reabsorbed (at most Tm) and excreted glucose and the plasma threshold Tm/GFR, ignoring splay.

E = \max(0,\; GFR\cdot P_G – T_m)

You give

	Meaning	Unit	Example
\text{GFR}	GFR (e.g. in mL/min)	m^3/s	125 mL/min
\text{Pglucose}	Plasma glucose in mg/dL	—	400
\text{Tm}	Transport maximum (e.g. in mg/min)	kg/s	375 mg/min

You get

	Meaning	Unit	Example
\text{filtered\_load}	Filtered load in mg/min	—	500
\text{reabsorbed}	Reabsorbed in mg/min	—	375
\text{excreted}	Excreted in mg/min	—	125
\text{threshold}	Plasma threshold in mg/dL	—	300

Values with a unit may be written in any unit of the right kind.

Applies when

GFR, Tm > 0
Pglucose >= 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

glucose titration: GFR = 125 mL/min, Pglucose = 400, Tm = 375 mg/min

filtered_load = 500

reabsorbed = 375

excreted = 125

threshold = 300

Run it

Terminal

$ human-physiology solve hp_glucose_titration "GFR=125 mL/min" Pglucose=400 "Tm=375 mg/min"
filtered_load = 500
reabsorbed = 375
excreted = 125
threshold = 300
  by: Glucose titration curve (ideal transport maximum)  (reabsorbed = min(filtered, Tm))
  GFR = 2.083333333e-06 m^3/s
  Pglucose = 400
  Tm = 6.25e-06 kg/s
  assumes: GFR, Tm > 0; Pglucose >= 0

import human_physiology_ai999x as m

a = m.solve("hp_glucose_titration", GFR="125 mL/min", Pglucose="400", Tm="375 mg/min")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Osmolar and free water clearance1 fig.hp_free_water_clearance

Osmolar clearance Cosm = Uosm V/Posm and free water clearance CH2O = V – Cosm (negative when the urine is concentrated).

C_{H_2O} = \dot V – \frac{U_{osm}\dot V}{P_{osm}}

You give

	Meaning	Unit	Example
V	Urine flow (e.g. in mL/min)	m^3/s	2 mL/min
\text{Uosm}	Urine osmolality in mOsm/kg	—	150
\text{Posm}	Plasma osmolality in mOsm/kg	—	300

You get

	Meaning	Unit	Example
\text{Cosm}	Osmolar clearance in mL/min	—	1
\text{CH2O}	Free water clearance in mL/min	—	1

Values with a unit may be written in any unit of the right kind.

Applies when

all inputs > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

free water clearance: V = 2 mL/min, Uosm = 150, Posm = 300

Cosm = 1

CH2O = 1

Run it

Terminal

$ human-physiology solve hp_free_water_clearance "V=2 mL/min" Uosm=150 Posm=300
Cosm = 1
CH2O = 1
  by: Osmolar and free water clearance  (CH2O = V (1 - Uosm/Posm))
  V = 3.333333333e-08 m^3/s
  Uosm = 150
  Posm = 300
  assumes: all inputs > 0

import human_physiology_ai999x as m

a = m.solve("hp_free_water_clearance", V="2 mL/min", Uosm="150", Posm="300")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Renal clearance of a substance (inulin GFR)1 fig.hp_renal_clearance

Clearance C = U V/P (equal to GFR for inulin) and the urinary excretion rate U V.

C_x = \frac{U_x \dot V}{P_x}

You give

	Meaning	Unit	Example
U	Urine concentration in mg/dL	—	30
V	Urine flow (e.g. in mL/min)	m^3/s	1 mL/min
P	Plasma concentration in mg/dL	—	0.25

You get

	Meaning	Unit	Example
C	Clearance in mL/min	—	120
\text{excretion}	Excretion rate in mg/min	—	0.3

Values with a unit may be written in any unit of the right kind.

Applies when

U >= 0
V, P > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

renal clearance: U = 30, V = 1 mL/min, P = 0.25

C = 120

excretion = 0.3

Run it

Terminal

$ human-physiology solve hp_renal_clearance U=30 "V=1 mL/min" P=0.25
C = 120
excretion = 0.3
  by: Renal clearance of a substance (inulin GFR)  (C = U V / P)
  U = 30
  V = 1.666666667e-08 m^3/s
  P = 0.25
  assumes: U >= 0; V, P > 0

import human_physiology_ai999x as m

a = m.solve("hp_renal_clearance", U="30", V="1 mL/min", P="0.25")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Renal plasma and blood flow from PAH clearance1 fig.hp_pah_renal_flow

Effective renal plasma flow = PAH clearance, true RPF = eRPF/E (PAH extraction) and RBF = RPF/(1 – Hct).

RBF = \frac{U_{PAH}\dot V / P_{PAH}}{E_{PAH}(1 – Hct)}

You give

	Meaning	Unit	Example
U_{\mathrm{PAH}}	Urine PAH concentration in mg/dL	—	600
V	Urine flow (e.g. in mL/min)	m^3/s	1 mL/min
P_{\mathrm{PAH}}	Plasma PAH concentration in mg/dL	—	1
\text{Hct}	Hematocrit (fraction)	—	0.45
E	PAH extraction ratio (optional; complete extraction, the effective flow, if omitted) optional	—	0.9

You get

	Meaning	Unit	Example
\text{eRPF}	Effective renal plasma flow in mL/min	—	600
\text{RPF}	Renal plasma flow in mL/min	—	666.6666667
\text{RBF}	Renal blood flow in mL/min	—	1212.121212

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

0 <= Hct < 1
0 < E <= 1

Outside these conditions it says what is wrong instead of returning a number.

Worked example

renal plasma flow: U_PAH = 600, V = 1 mL/min, P_PAH = 1, Hct = 0.45, E = 0.9

eRPF = 600

RPF = 666.6666667

RBF = 1212.121212

Run it

Terminal

$ human-physiology solve hp_pah_renal_flow U_PAH=600 "V=1 mL/min" P_PAH=1 Hct=0.45 E=0.9
eRPF = 600
RPF = 666.6666667
RBF = 1212.121212
  by: Renal plasma and blood flow from PAH clearance  (eRPF = U V/P; RPF = eRPF/E; RBF = RPF/(1 - Hct))
  U_PAH = 600
  V = 1.666666667e-08 m^3/s
  P_PAH = 1
  Hct = 0.45
  E = 0.9
  assumes: 0 <= Hct < 1; 0 < E <= 1

import human_physiology_ai999x as m

a = m.solve("hp_pah_renal_flow", U_PAH="600", V="1 mL/min", P_PAH="1", Hct="0.45", E="0.9")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Respiratory 27

Alveolar gas equation1 fig.hp_alveolar_gas_equation

Inspired PO2 = FiO2 (PB – PH2O) and alveolar PO2 by the simplified (PiO2 – PaCO2/R) and full (nitrogen-balance) alveolar gas equation.

P_AO_2 = F_iO_2(P_B – P_{H_2O}) – \frac{P_aCO_2}{R}\left[1 – F_iO_2(1 – R)\right]

You give

	Meaning	Unit	Example
\text{FiO2}	Inspired O2 fraction	—	0.21
\text{Patm}	Barometric pressure (e.g. in mmHg)	Pa	760 mmHg
\text{PH2O}	Water vapour pressure (e.g. in mmHg)	Pa	47 mmHg
\text{PaCO2}	Arterial PCO2 (e.g. in mmHg)	Pa	40 mmHg
\text{RQ}	Respiratory quotient	—	0.8

You get

	Meaning	Unit	Example
\text{PiO2}	Inspired PO2 in mmHg	—	149.73
\text{PAO2}	Alveolar PO2, simplified equation, in mmHg	—	99.73
\text{PAO2\_full}	Alveolar PO2, full equation, in mmHg	—	101.83

Values with a unit may be written in any unit of the right kind.

Applies when

0 < FiO2 <= 1
Patm > PH2O
0 < RQ <= 1.5

Outside these conditions it says what is wrong instead of returning a number.

Worked example

alveolar gas equation: FiO2 = 0.21, Patm = 760 mmHg, PH2O = 47 mmHg, PaCO2 = 40 mmHg, RQ = 0.8

PiO2 = 149.73

PAO2 = 99.73

PAO2_full = 101.83

Run it

Terminal

$ human-physiology solve hp_alveolar_gas_equation FiO2=0.21 "Patm=760 mmHg" "PH2O=47 mmHg" "PaCO2=40 mmHg" RQ=0.8
PiO2 = 149.73
PAO2 = 99.73
PAO2_full = 101.83
  by: Alveolar gas equation  (PAO2 = PiO2 - PaCO2/R (+ PaCO2 FiO2 (1 - R)/R))
  FiO2 = 0.21
  Patm = 101325.0144 Pa
  PH2O = 6266.152209 Pa
  PaCO2 = 5332.895497 Pa
  RQ = 0.8
  assumes: 0 < FiO2 <= 1; Patm > PH2O; 0 < RQ <= 1.5

import human_physiology_ai999x as m

a = m.solve("hp_alveolar_gas_equation", FiO2="0.21", Patm="760 mmHg", PH2O="47 mmHg", PaCO2="40 mmHg", RQ="0.8")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Alveolar ventilation from CO2 output and arterial PCO22 fig.hp_alveolar_ventilation_co2

Alveolar ventilation equation VA (BTPS) = VCO2 (STPD) x 863/PaCO2, with K = 760 x 310.15/273.15 mmHg.

\dot V_A = \frac{\dot V_{CO_2} \times 0.863}{P_aCO_2}

You give

	Meaning	Unit	Example
\text{VCO2}	CO2 output (e.g. in mL/min) (STPD)	m^3/s	200 mL/min
\text{PaCO2}	Arterial (alveolar) PCO2 (e.g. in mmHg)	Pa	40 mmHg

You get

	Meaning	Unit	Example
\text{VA}	Alveolar ventilation in L/min (BTPS)	—	4.314735493

Values with a unit may be written in any unit of the right kind.

Applies when

VCO2, PaCO2 > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

alveolar ventilation from CO2 output: VCO2 = 200 mL/min, PaCO2 = 40 mmHg

VA = 4.314735493

Run it

Terminal

$ human-physiology solve hp_alveolar_ventilation_co2 "VCO2=200 mL/min" "PaCO2=40 mmHg"
VA = 4.314735493
  by: Alveolar ventilation from CO2 output and arterial PCO2  (VA = VCO2 K / PaCO2, K = 760 x 310.15/273.15)
  VCO2 = 3.333333333e-06 m^3/s
  PaCO2 = 5332.895497 Pa
  assumes: VCO2, PaCO2 > 0

import human_physiology_ai999x as m

a = m.solve("hp_alveolar_ventilation_co2", VCO2="200 mL/min", PaCO2="40 mmHg")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Alveolar-arterial oxygen gradient, a/A ratio and P/F ratio1 fig.hp_aa_gradient

PAO2 from the simplified alveolar gas equation, the A-a gradient PAO2 – PaO2, the a/A ratio and PaO2/FiO2.

A\!-\!a = P_AO_2 – P_aO_2

You give

	Meaning	Unit	Example
\text{FiO2}	Inspired O2 fraction	—	0.21
\text{Patm}	Barometric pressure (e.g. in mmHg)	Pa	760 mmHg
\text{PH2O}	Water vapour pressure (e.g. in mmHg)	Pa	47 mmHg
\text{PaCO2}	Arterial PCO2 (e.g. in mmHg)	Pa	40 mmHg
\text{PaO2}	Arterial PO2 (e.g. in mmHg)	Pa	90 mmHg
\text{RQ}	Respiratory quotient	—	0.8

You get

	Meaning	Unit	Example
\text{PAO2}	Alveolar PO2 in mmHg	—	99.73
\text{Aa\_gradient}	A-a gradient in mmHg	—	9.73
a_{A,\mathrm{ratio}}	PaO2/PAO2	—	0.9024365788
\text{PF\_ratio}	PaO2/FiO2 in mmHg	—	428.5714286

Values with a unit may be written in any unit of the right kind.

Applies when

0 < FiO2 <= 1
computed PAO2 > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

A-a gradient: FiO2 = 0.21, Patm = 760 mmHg, PH2O = 47 mmHg, PaCO2 = 40 mmHg, PaO2 = 90 mmHg, RQ = 0.8

PAO2 = 99.73

Aa_gradient = 9.73

a_A_ratio = 0.9024365788

PF_ratio = 428.5714286

Run it

Terminal

$ human-physiology solve hp_aa_gradient FiO2=0.21 "Patm=760 mmHg" "PH2O=47 mmHg" "PaCO2=40 mmHg" "PaO2=90 mmHg" RQ=0.8
PAO2 = 99.73
Aa_gradient = 9.73
a_A_ratio = 0.9024365788
PF_ratio = 428.5714286
  by: Alveolar-arterial oxygen gradient, a/A ratio and P/F ratio  (A-a = FiO2 (PB - PH2O) - PaCO2/R - PaO2)
  FiO2 = 0.21
  Patm = 101325.0144 Pa
  PH2O = 6266.152209 Pa
  PaCO2 = 5332.895497 Pa
  PaO2 = 11999.01487 Pa
  RQ = 0.8
  assumes: 0 < FiO2 <= 1; computed PAO2 > 0

import human_physiology_ai999x as m

a = m.solve("hp_aa_gradient", FiO2="0.21", Patm="760 mmHg", PH2O="47 mmHg", PaCO2="40 mmHg", PaO2="90 mmHg", RQ="0.8")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Arterial oxygen content1 fig.hp_oxygen_content

CaO2 = 1.34 x Hb x SaO2 + 0.003 x PaO2 (hemoglobin-bound plus dissolved oxygen).

C_aO_2 = 1.34\,Hb\,S_aO_2 + 0.003\,P_aO_2

You give

	Meaning	Unit	Example
\text{Hb}	Hemoglobin in g/dL	—	15
\text{SaO2}	Arterial saturation	—	0.98
\text{PaO2}	Arterial PO2 (e.g. in mmHg)	Pa	100 mmHg

You get

	Meaning	Unit	Example
\text{CaO2}	Oxygen content in mL O2/dL	—	19.998
\text{bound}	Hemoglobin-bound O2 in mL/dL	—	19.698
\text{dissolved}	Dissolved O2 in mL/dL	—	0.3

Values with a unit may be written in any unit of the right kind.

Applies when

Hb > 0
0 <= SaO2 <= 1

Outside these conditions it says what is wrong instead of returning a number.

Worked example

oxygen content: Hb = 15, SaO2 = 0.98, PaO2 = 100 mmHg

CaO2 = 19.998

bound = 19.698

dissolved = 0.3

Run it

Terminal

$ human-physiology solve hp_oxygen_content Hb=15 SaO2=0.98 "PaO2=100 mmHg"
CaO2 = 19.998
bound = 19.698
dissolved = 0.3
  by: Arterial oxygen content  (1.34 Hb S + 0.003 PO2)
  Hb = 15
  SaO2 = 0.98
  PaO2 = 13332.23874 Pa
  assumes: Hb > 0; 0 <= SaO2 <= 1

import human_physiology_ai999x as m

a = m.solve("hp_oxygen_content", Hb="15", SaO2="0.98", PaO2="100 mmHg")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

ATPS to BTPS and STPD gas volume conversion1 fig.hp_gas_volume_conversion

A spirometer volume measured at ambient temperature and pressure, saturated (ATPS), converted to body (BTPS) and standard dry (STPD) conditions.

V_{BTPS} = V\,\frac{310.15}{T}\,\frac{P_B – P_{H_2O}}{P_B – 47}

You give

	Meaning	Unit	Example
V	Measured volume (e.g. in L)	m^3	1 L
T	Spirometer temperature (e.g. in degC)	K	22 degC
\text{Patm}	Barometric pressure (e.g. in mmHg)	Pa	760 mmHg
\text{PH2O}	Water vapour pressure at T (e.g. in mmHg)	Pa	19.8 mmHg

You get

	Meaning	Unit	Example
V_{\mathrm{BTPS}}	Volume at BTPS in L	—	1.090909061
V_{\mathrm{STPD}}	Volume at STPD in L	—	0.9013509188

Values with a unit may be written in any unit of the right kind.

Applies when

Patm > PH2O
Patm > 47 mmHg

Outside these conditions it says what is wrong instead of returning a number.

Worked example

gas volume conversion: V = 1 L, T = 22 degC, Patm = 760 mmHg, PH2O = 19.8 mmHg

V_BTPS = 1.090909061

V_STPD = 0.9013509188

Run it

Terminal

$ human-physiology solve hp_gas_volume_conversion "V=1 L" "T=22 degC" "Patm=760 mmHg" "PH2O=19.8 mmHg"
V_BTPS = 1.090909061
V_STPD = 0.9013509188
  by: ATPS to BTPS and STPD gas volume conversion  (ideal gas law on the dry-gas partial pressures)
  V = 0.001 m^3
  T = 295.15 K
  Patm = 101325.0144 Pa
  PH2O = 2639.783271 Pa
  assumes: Patm > PH2O; Patm > 47 mmHg

import human_physiology_ai999x as m

a = m.solve("hp_gas_volume_conversion", V="1 L", T="22 degC", Patm="760 mmHg", PH2O="19.8 mmHg")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Barometric pressure and inspired PO2 at altitude1 fig.hp_barometric_altitude

Standard-atmosphere barometric pressure P = 101325 (1 – 0.0065 h/288.15)^(g M/(R L)) and the humidified inspired PO2.

P = P_0\left(1 – \frac{L h}{T_0}\right)^{gM/(RL)}

You give

	Meaning	Unit	Example
h	Altitude (e.g. in m)	m	3000 m
\text{FiO2}	Inspired oxygen fraction (optional; dry air if omitted) optional	—

You get

	Meaning	Unit	Example
\text{Patm}	Barometric pressure in mmHg	—	525.8605769
\text{PiO2}	Humidified inspired PO2 in mmHg	—	100.3212909
\text{Patm\_kPa}	Barometric pressure in kPa	—	70.10898756

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

0 <= h <= 11000 m

Outside these conditions it says what is wrong instead of returning a number.

Worked example

barometric pressure at altitude: h = 3000 m

Patm = 525.8605769

PiO2 = 100.3212909

Patm_kPa = 70.10898756

Run it

Terminal

$ human-physiology solve hp_barometric_altitude "h=3000 m"
Patm = 525.8605769
PiO2 = 100.3212909
Patm_kPa = 70.10898756
  by: Barometric pressure and inspired PO2 at altitude  (ICAO standard troposphere (lapse 6.5 K/km))
  h = 3000 m
  assumes: 0 <= h <= 11000 m

import human_physiology_ai999x as m

a = m.solve("hp_barometric_altitude", h="3000 m")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Bicarbonate from pH and PCO2 (Henderson-Hasselbalch)1 fig.hp_bicarbonate_from_ph

HCO3 = 0.03 PaCO2 x 10^(pH – 6.1).

[HCO_3^-] = 0.03\,P_{CO_2}\,10^{pH – 6.1}

You give

	Meaning	Unit	Example
\text{pH}	PH	—	7.4
\text{PaCO2}	PCO2 (e.g. in mmHg)	Pa	40 mmHg

You get

	Meaning	Unit	Example
\text{HCO3}	Bicarbonate in mmol/L	—	23.94314778

Values with a unit may be written in any unit of the right kind.

Applies when

6.5 <= pH <= 8.0
PaCO2 > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

bicarbonate from pH: pH = 7.4, PaCO2 = 40 mmHg

HCO3 = 23.94314778

Run it

Terminal

$ human-physiology solve hp_bicarbonate_from_ph pH=7.4 "PaCO2=40 mmHg"
HCO3 = 23.94314778
  by: Bicarbonate from pH and PCO2 (Henderson-Hasselbalch)  (Henderson-Hasselbalch solved for bicarbonate)
  pH = 7.4
  PaCO2 = 5332.895497 Pa
  assumes: 6.5 <= pH <= 8.0; PaCO2 > 0

import human_physiology_ai999x as m

a = m.solve("hp_bicarbonate_from_ph", pH="7.4", PaCO2="40 mmHg")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Blood pH by the Henderson-Hasselbalch equation2 fig.hp_ph_henderson_hasselbalch

pH = 6.1 + log10(HCO3/(0.03 PaCO2)) and the hydrogen-ion concentration in nmol/L.

pH = 6.1 + \log_{10}\frac{[HCO_3^-]}{0.03\,P_{CO_2}}

You give

	Meaning	Unit	Example
\text{HCO3}	Bicarbonate (e.g. in mmol/L)	mol/m^3	24 mmol/L
\text{PaCO2}	PCO2 (e.g. in mmHg)	Pa	40 mmHg

You get

	Meaning	Unit	Example
\text{pH}	PH	—	7.401029996
H	Hydrogen-ion concentration in nmol/L	—	39.71641174

Values with a unit may be written in any unit of the right kind.

Applies when

HCO3, PaCO2 > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

blood pH: HCO3 = 24 mmol/L, PaCO2 = 40 mmHg

pH = 7.401029996

H = 39.71641174

Run it

Terminal

$ human-physiology solve hp_ph_henderson_hasselbalch "HCO3=24 mmol/L" "PaCO2=40 mmHg"
pH = 7.401029996
H = 39.71641174
  by: Blood pH by the Henderson-Hasselbalch equation  (pK' 6.1, CO2 solubility 0.03 mmol/L per mmHg)
  HCO3 = 24 mol/m^3
  PaCO2 = 5332.895497 Pa
  assumes: HCO3, PaCO2 > 0

import human_physiology_ai999x as m

a = m.solve("hp_ph_henderson_hasselbalch", HCO3="24 mmol/L", PaCO2="40 mmHg")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

FEV1/FVC ratio2 fig.hp_fev1_fvc_ratio

The fraction of the forced vital capacity exhaled in the first second.

\frac{FEV_1}{FVC}

You give

	Meaning	Unit	Example
\text{FEV1}	Forced expiratory volume in the first second (e.g. in L)	m^3	3.2 L
\text{FVC}	Forced vital capacity (e.g. in L)	m^3	4 L

You get

	Meaning	Unit	Example
\text{ratio}	FEV1/FVC	—	0.8
\text{percent}	FEV1/FVC in %	—	80

Values with a unit may be written in any unit of the right kind.

Applies when

0 < FEV1 <= FVC

Outside these conditions it says what is wrong instead of returning a number.

Worked example

FEV1/FVC ratio: FEV1 = 3.2 L, FVC = 4 L

ratio = 0.8

percent = 80

Run it

Terminal

$ human-physiology solve hp_fev1_fvc_ratio "FEV1=3.2 L" "FVC=4 L"
ratio = 0.8
percent = 80
  by: FEV1/FVC ratio  (FEV1 / FVC)
  FEV1 = 0.0032 m^3
  FVC = 0.004 m^3
  assumes: 0 < FEV1 <= FVC

import human_physiology_ai999x as m

a = m.solve("hp_fev1_fvc_ratio", FEV1="3.2 L", FVC="4 L")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Fick's law of diffusion across a membrane1 fig.hp_fick_diffusion

Steady diffusive flux J = D A dC/T across a barrier of thickness T, and the diffusion time T^2/(2D).

J = D A \frac{\Delta C}{T},\quad t = \frac{T^2}{2D}

You give

	Meaning	Unit	Example
D	Diffusion coefficient (e.g. in cm²/s)	m^2/s	1.5e-05 cm^2/s
A	Membrane area (e.g. in cm²)	m^2	2 cm^2
\text{dC}	Concentration difference (e.g. in mmol/L)	mol/m^3	0.5 mmol/L
T	Membrane thickness (e.g. in um)	m	2 um

You get

	Meaning	Unit	Example
J	Diffusion rate in mmol/s	—	7.5e-05
t_{\mathrm{diffusion}}	Characteristic diffusion time across the membrane in s	—	0.001333333333

Values with a unit may be written in any unit of the right kind.

Applies when

D, A, T > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Fick diffusion: D = 1.5e-05 cm^2/s, A = 2 cm^2, dC = 0.5 mmol/L, T = 2 um

J = 7.5e-05

t_diffusion = 0.001333333333

Run it

Terminal

$ human-physiology solve hp_fick_diffusion "D=1.5e-05 cm^2/s" "A=2 cm^2" "dC=0.5 mmol/L" "T=2 um"
J = 7.5e-05
t_diffusion = 0.001333333333
  by: Fick's law of diffusion across a membrane  (J = D A dC / T (units converted to cm))
  D = 1.5e-09 m^2/s
  A = 0.0002 m^2
  dC = 0.5 mol/m^3
  T = 2e-06 m
  assumes: D, A, T > 0

import human_physiology_ai999x as m

a = m.solve("hp_fick_diffusion", D="1.5e-05 cm^2/s", A="2 cm^2", dC="0.5 mmol/L", T="2 um")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Hill coefficient and P50 from two points of the dissociation curve1 fig.hp_hill_fit

The Hill coefficient n and P50 of the curve through two measured (PO2, saturation) points (Hill plot slope).

n = \frac{\ln\frac{S_2/(1-S_2)}{S_1/(1-S_1)}}{\ln(P_2/P_1)}

You give

	Meaning	Unit	Example
\text{P1}	First PO2 (e.g. in mmHg)	Pa	20 mmHg
\text{S1}	Saturation at P1	—	0.3121240186
\text{P2}	Second PO2 (e.g. in mmHg)	Pa	60 mmHg
\text{S2}	Saturation at P2	—	0.8980776441

You get

	Meaning	Unit	Example
n	Hill coefficient	—	2.7
\text{P50}	P50 in mmHg	—	26.8

Values with a unit may be written in any unit of the right kind.

Applies when

0 < S < 1
saturation increases with PO2

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Hill fit: P1 = 20 mmHg, S1 = 0.312124018611, P2 = 60 mmHg, S2 = 0.898077644148

n = 2.7

P50 = 26.8

Run it

Terminal

$ human-physiology solve hp_hill_fit "P1=20 mmHg" S1=0.3121240186 "P2=60 mmHg" S2=0.8980776441
n = 2.7
P50 = 26.8
  by: Hill coefficient and P50 from two points of the dissociation curve  (slope of log(S/(1-S)) against log PO2)
  P1 = 2666.447748 Pa
  S1 = 0.3121240186
  P2 = 7999.343245 Pa
  S2 = 0.8980776441
  assumes: 0 < S < 1; saturation increases with PO2

import human_physiology_ai999x as m

a = m.solve("hp_hill_fit", P1="20 mmHg", S1="0.3121240186", P2="60 mmHg", S2="0.8980776441")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Laplace pressure of an alveolus2 fig.hp_alveolar_laplace_pressure

Pressure needed to keep a spherical alveolus open, P = 2 gamma/r (why surfactant matters).

P = \frac{2\gamma}{r}

You give

	Meaning	Unit	Example
\gamma	Surface tension (e.g. in mN/m)	N/m	25 mN/m
r	Alveolar radius (e.g. in um)	m	100 um

You get

	Meaning	Unit	Example
P	Collapsing pressure in Pa	—	500
P_{\mathrm{cmH2O}}	Collapsing pressure in cmH2O	—	5.098581065

Values with a unit may be written in any unit of the right kind.

Applies when

gamma, r > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

alveolar Laplace pressure: gamma = 25 mN/m, r = 100 um

P = 500

P_cmH2O = 5.098581065

Run it

Terminal

$ human-physiology solve hp_alveolar_laplace_pressure "gamma=25 mN/m" "r=100 um"
P = 500
P_cmH2O = 5.098581065
  by: Laplace pressure of an alveolus  (P = 2 gamma / r)
  gamma = 0.025 N/m
  r = 0.0001 m
  assumes: gamma, r > 0

import human_physiology_ai999x as m

a = m.solve("hp_alveolar_laplace_pressure", gamma="25 mN/m", r="100 um")
print(a.values or (a.value, a.unit))

Figures

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Lung capacities from the four lung volumes1 fig.hp_lung_capacities

Inspiratory capacity, functional residual capacity, vital capacity and total lung capacity from TV, IRV, ERV and RV.

TLC = IRV + TV + ERV + RV

You give

	Meaning	Unit	Example
\text{TV}	Tidal volume (e.g. in mL)	m^3	500 mL
\text{IRV}	Inspiratory reserve volume (e.g. in mL)	m^3	3000 mL
\text{ERV}	Expiratory reserve volume (e.g. in mL)	m^3	1100 mL
\text{RV}	Residual volume (e.g. in mL)	m^3	1200 mL

You get

	Meaning	Unit	Example
\text{IC}	Inspiratory capacity in mL	—	3500
\text{FRC}	Functional residual capacity in mL	—	2300
\text{VC}	Vital capacity in mL	—	4600
\text{TLC}	Total lung capacity in mL	—	5800

Values with a unit may be written in any unit of the right kind.

Applies when

TV, RV > 0
IRV, ERV >= 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

lung capacities: TV = 500 mL, IRV = 3000 mL, ERV = 1100 mL, RV = 1200 mL

IC = 3500

FRC = 2300

VC = 4600

TLC = 5800

Run it

Terminal

$ human-physiology solve hp_lung_capacities "TV=500 mL" "IRV=3000 mL" "ERV=1100 mL" "RV=1200 mL"
IC = 3500
FRC = 2300
VC = 4600
TLC = 5800
  by: Lung capacities from the four lung volumes  (IC = TV + IRV; FRC = ERV + RV; VC = IRV + TV + ERV; TLC = VC + RV)
  TV = 0.0005 m^3
  IRV = 0.003 m^3
  ERV = 0.0011 m^3
  RV = 0.0012 m^3
  assumes: TV, RV > 0; IRV, ERV >= 0

import human_physiology_ai999x as m

a = m.solve("hp_lung_capacities", TV="500 mL", IRV="3000 mL", ERV="1100 mL", RV="1200 mL")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Minute and alveolar ventilation1 fig.hp_minute_ventilation

Minute ventilation VE = VT x RR, alveolar ventilation VA = (VT – VD) x RR and the dead-space fraction VD/VT.

\dot V_A = (V_T – V_D) f

You give

	Meaning	Unit	Example
\text{VT}	Tidal volume (e.g. in mL)	m^3	500 mL
\text{RR}	Respiratory rate in breaths/min	—	12
\text{VD}	Dead space (e.g. in mL)	m^3	150 mL

You get

	Meaning	Unit	Example
\text{VE}	Minute ventilation in L/min	—	6
\text{VA}	Alveolar ventilation in L/min	—	4.2
\text{VD\_VT}	Dead space fraction	—	0.3

Values with a unit may be written in any unit of the right kind.

Applies when

0 <= VD < VT
RR > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

minute ventilation: VT = 500 mL, RR = 12, VD = 150 mL

VE = 6

VA = 4.2

VD_VT = 0.3

Run it

Terminal

$ human-physiology solve hp_minute_ventilation "VT=500 mL" RR=12 "VD=150 mL"
VE = 6
VA = 4.2
VD_VT = 0.3
  by: Minute and alveolar ventilation  (VE = VT f; VA = (VT - VD) f)
  VT = 0.0005 m^3
  RR = 12
  VD = 0.00015 m^3
  assumes: 0 <= VD < VT; RR > 0

import human_physiology_ai999x as m

a = m.solve("hp_minute_ventilation", VT="500 mL", RR="12", VD="150 mL")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Oxygen consumption and extraction ratio (reverse Fick)1 fig.hp_oxygen_consumption

VO2 = CO (CaO2 – CvO2) x 10 and the oxygen extraction ratio (CaO2 – CvO2)/CaO2.

\dot V_{O_2} = CO\,(C_aO_2 – C_{\bar v}O_2)

You give

	Meaning	Unit	Example
\text{CO}	Cardiac output (e.g. in L/min)	m^3/s	5 L/min
\text{Hb}	Hemoglobin in g/dL	—	15
\text{SaO2}	Arterial saturation	—	0.98
\text{SvO2}	Mixed venous saturation	—	0.75
\text{PaO2}	Arterial PO2 (e.g. in mmHg)	Pa	100 mmHg
\text{PvO2}	Mixed venous PO2 (e.g. in mmHg)	Pa	40 mmHg

You get

	Meaning	Unit	Example
\text{VO2}	Oxygen consumption in mL/min	—	240.15
\text{O2ER}	Oxygen extraction ratio	—	0.2401740174
\text{CaO2}	Arterial content in mL/dL	—	19.998
\text{CvO2}	Venous content in mL/dL	—	15.195

Values with a unit may be written in any unit of the right kind.

Applies when

CvO2 < CaO2

Outside these conditions it says what is wrong instead of returning a number.

Worked example

oxygen consumption: CO = 5 L/min, Hb = 15, SaO2 = 0.98, SvO2 = 0.75, PaO2 = 100 mmHg, PvO2 = 40 mmHg

VO2 = 240.15

O2ER = 0.2401740174

CaO2 = 19.998

CvO2 = 15.195

Run it

Terminal

$ human-physiology solve hp_oxygen_consumption "CO=5 L/min" Hb=15 SaO2=0.98 SvO2=0.75 "PaO2=100 mmHg" "PvO2=40 mmHg"
VO2 = 240.15
O2ER = 0.2401740174
CaO2 = 19.998
CvO2 = 15.195
  by: Oxygen consumption and extraction ratio (reverse Fick)  (VO2 = 10 CO (CaO2 - CvO2))
  CO = 8.333333333e-05 m^3/s
  Hb = 15
  SaO2 = 0.98
  SvO2 = 0.75
  PaO2 = 13332.23874 Pa
  PvO2 = 5332.895497 Pa
  assumes: CvO2 < CaO2

import human_physiology_ai999x as m

a = m.solve("hp_oxygen_consumption", CO="5 L/min", Hb="15", SaO2="0.98", SvO2="0.75", PaO2="100 mmHg", PvO2="40 mmHg")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Oxyhemoglobin saturation by the Hill equation1 fig.hp_hill_saturation

SO2 = PO2^n/(PO2^n + P50^n) for a Hill coefficient n and half-saturation pressure P50.

S = \frac{P^n}{P^n + P_{50}^n}

You give

	Meaning	Unit	Example
\text{PO2}	Partial pressure of O2 (e.g. in mmHg)	Pa	40 mmHg
\text{P50}	P50 (e.g. in mmHg)	Pa	26.8 mmHg
n	Hill coefficient	—	2.7

You get

	Meaning	Unit	Example
\text{SO2}	Fractional saturation	—	0.7467379184

Values with a unit may be written in any unit of the right kind.

Applies when

PO2 >= 0
P50, n > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Hill saturation: PO2 = 40 mmHg, P50 = 26.8 mmHg, n = 2.7

SO2 = 0.7467379184

Run it

Terminal

$ human-physiology solve hp_hill_saturation "PO2=40 mmHg" "P50=26.8 mmHg" n=2.7
SO2 = 0.7467379184
  by: Oxyhemoglobin saturation by the Hill equation  (Hill equation)
  PO2 = 5332.895497 Pa
  P50 = 3573.039983 Pa
  n = 2.7
  assumes: PO2 >= 0; P50, n > 0

import human_physiology_ai999x as m

a = m.solve("hp_hill_saturation", PO2="40 mmHg", P50="26.8 mmHg", n="2.7")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

PCO2 from pH and bicarbonate (Henderson-Hasselbalch)1 fig.hp_pco2_from_ph

PaCO2 = HCO3/(0.03 x 10^(pH – 6.1)).

P_{CO_2} = \frac{[HCO_3^-]}{0.03\cdot 10^{pH – 6.1}}

You give

	Meaning	Unit	Example
\text{pH}	PH	—	7.4
\text{HCO3}	Bicarbonate (e.g. in mmol/L)	mol/m^3	24 mmol/L

You get

	Meaning	Unit	Example
\text{PaCO2}	PCO2 in mmHg	—	40.09497869

Values with a unit may be written in any unit of the right kind.

Applies when

6.5 <= pH <= 8.0
HCO3 > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

PCO2 from pH: pH = 7.4, HCO3 = 24 mmol/L

PaCO2 = 40.09497869

Run it

Terminal

$ human-physiology solve hp_pco2_from_ph pH=7.4 "HCO3=24 mmol/L"
PaCO2 = 40.09497869
  by: PCO2 from pH and bicarbonate (Henderson-Hasselbalch)  (Henderson-Hasselbalch solved for PCO2)
  pH = 7.4
  HCO3 = 24 mol/m^3
  assumes: 6.5 <= pH <= 8.0; HCO3 > 0

import human_physiology_ai999x as m

a = m.solve("hp_pco2_from_ph", pH="7.4", HCO3="24 mmol/L")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Physiological dead space by the Bohr (Enghoff) equation1 fig.hp_bohr_dead_space

VD/VT = (PaCO2 – PECO2)/PaCO2 and the dead-space volume, with arterial PCO2 standing for alveolar (Enghoff).

\frac{V_D}{V_T} = \frac{P_aCO_2 – P_ECO_2}{P_aCO_2}

You give

	Meaning	Unit	Example
\text{PaCO2}	Arterial PCO2 (e.g. in mmHg)	Pa	40 mmHg
\text{PECO2}	Mixed expired PCO2 (e.g. in mmHg)	Pa	28 mmHg
\text{VT}	Tidal volume (e.g. in mL)	m^3	500 mL

You get

	Meaning	Unit	Example
\text{VD\_VT}	Dead space fraction	—	0.3
\text{VD}	Dead space volume in mL	—	150

Values with a unit may be written in any unit of the right kind.

Applies when

0 <= PECO2 < PaCO2

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Bohr dead space: PaCO2 = 40 mmHg, PECO2 = 28 mmHg, VT = 500 mL

VD_VT = 0.3

VD = 150

Run it

Terminal

$ human-physiology solve hp_bohr_dead_space "PaCO2=40 mmHg" "PECO2=28 mmHg" "VT=500 mL"
VD_VT = 0.3
VD = 150
  by: Physiological dead space by the Bohr (Enghoff) equation  (Bohr CO2 mass balance)
  PaCO2 = 5332.895497 Pa
  PECO2 = 3733.026848 Pa
  VT = 0.0005 m^3
  assumes: 0 <= PECO2 < PaCO2

import human_physiology_ai999x as m

a = m.solve("hp_bohr_dead_space", PaCO2="40 mmHg", PECO2="28 mmHg", VT="500 mL")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Physiological shunt fraction (shunt equation)1 fig.hp_shunt_fraction

Qs/Qt = (Cc'O2 – CaO2)/(Cc'O2 – CvO2) from end-capillary, arterial and mixed venous O2 contents.

\frac{Q_s}{Q_t} = \frac{C_{c'}O_2 – C_aO_2}{C_{c'}O_2 – C_{\bar v}O_2}

You give

	Meaning	Unit	Example
\text{CcO2}	End-capillary O2 content in mL/dL	—	21
\text{CaO2}	Arterial O2 content in mL/dL	—	20
\text{CvO2}	Mixed venous O2 content in mL/dL	—	15

You get

	Meaning	Unit	Example
\text{Qs\_Qt}	Shunt fraction	—	0.1666666667

Applies when

CcO2 >= CaO2 > CvO2

Outside these conditions it says what is wrong instead of returning a number.

Worked example

shunt fraction: CcO2 = 21, CaO2 = 20, CvO2 = 15

Qs_Qt = 0.1666666667

Run it

Terminal

$ human-physiology solve hp_shunt_fraction CcO2=21 CaO2=20 CvO2=15
Qs_Qt = 0.1666666667
  by: Physiological shunt fraction (shunt equation)  (mixing balance of shunted and capillary blood)
  CcO2 = 21
  CaO2 = 20
  CvO2 = 15
  assumes: CcO2 >= CaO2 > CvO2

import human_physiology_ai999x as m

a = m.solve("hp_shunt_fraction", CcO2="21", CaO2="20", CvO2="15")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

PO2 at a given saturation by the Hill equation1 fig.hp_hill_po2

Inverse Hill equation PO2 = P50 (S/(1 – S))^(1/n).

P = P_{50}\left(\frac{S}{1-S}\right)^{1/n}

You give

	Meaning	Unit	Example
\text{SO2}	Fractional saturation	—	0.75
\text{P50}	P50 (e.g. in mmHg)	Pa	26.8 mmHg
n	Hill coefficient	—	2.7

You get

	Meaning	Unit	Example
\text{PO2}	Partial pressure of O2 in mmHg	—	40.25745998

Values with a unit may be written in any unit of the right kind.

Applies when

0 < SO2 < 1
P50, n > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Hill PO2: SO2 = 0.75, P50 = 26.8 mmHg, n = 2.7

PO2 = 40.25745998

Run it

Terminal

$ human-physiology solve hp_hill_po2 SO2=0.75 "P50=26.8 mmHg" n=2.7
PO2 = 40.25745998
  by: PO2 at a given saturation by the Hill equation  (PO2 = P50 (S/(1-S))^(1/n))
  SO2 = 0.75
  P50 = 3573.039983 Pa
  n = 2.7
  assumes: 0 < SO2 < 1; P50, n > 0

import human_physiology_ai999x as m

a = m.solve("hp_hill_po2", SO2="0.75", P50="26.8 mmHg", n="2.7")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

PO2 at a given saturation by the Severinghaus equation1 fig.hp_severinghaus_po2

Inverse of the Severinghaus curve: the real root of P^3 + 150 P = 23400 S/(1 – S).

P^3 + 150P = \frac{23400\,S}{1-S}

You give

	Meaning	Unit	Example
\text{SO2}	Fractional saturation	—	0.7494646681

You get

	Meaning	Unit	Example
\text{PO2}	Partial pressure of O2 in mmHg	—	40

Applies when

0 < SO2 < 1

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Severinghaus PO2: SO2 = 0.749464668094

PO2 = 40

Run it

Terminal

$ human-physiology solve hp_severinghaus_po2 SO2=0.7494646681
PO2 = 40
  by: PO2 at a given saturation by the Severinghaus equation  (Cardano root of the cubic, Newton polished)
  SO2 = 0.7494646681
  assumes: 0 < SO2 < 1

import human_physiology_ai999x as m

a = m.solve("hp_severinghaus_po2", SO2="0.7494646681")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Respiratory time constant and passive exhalation1 fig.hp_respiratory_time_constant

tau = R C, the fraction of the volume exhaled after time t, 1 – e^(-t/tau), and 3 tau (about 95 %).

\tau = R C,\quad V(t) = V_0 e^{-t/\tau}

You give

	Meaning	Unit	Example
R	Airway resistance in cmH2O s/L	—	5
C	Compliance in mL/cmH2O	—	50
t	Time (e.g. in s) (optional; needed for the fraction exhaled) optional	s	0.5 s

You get

	Meaning	Unit	Example
\tau	Time constant in s	—	0.25
\text{fraction\_exhaled}	Fraction exhaled at time t	—	0.8646647168
t_{95}	Three time constants in s	—	0.75

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

R, C > 0
t >= 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

respiratory time constant: R = 5, C = 50, t = 0.5 s

tau = 0.25

t_95 = 0.75

fraction_exhaled = 0.8646647168

Run it

Terminal

$ human-physiology solve hp_respiratory_time_constant R=5 C=50 "t=0.5 s"
tau = 0.25
t_95 = 0.75
fraction_exhaled = 0.8646647168
  by: Respiratory time constant and passive exhalation  (single-compartment passive emptying)
  R = 5
  C = 50
  t = 0.5 s
  assumes: R, C > 0; t >= 0

import human_physiology_ai999x as m

a = m.solve("hp_respiratory_time_constant", R="5", C="50", t="0.5 s")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Saturation by the Severinghaus (1979) dissociation equation1 fig.hp_severinghaus_saturation

Standard human oxyhemoglobin curve S = 1/(23400/(PO2^3 + 150 PO2) + 1) (37 degC, pH 7.40, PCO2 40 mmHg).

S = \left(\frac{23400}{P^3 + 150P} + 1\right)^{-1}

You give

	Meaning	Unit	Example
\text{PO2}	Partial pressure of O2 (e.g. in mmHg)	Pa	40 mmHg

You get

	Meaning	Unit	Example
\text{SO2}	Fractional saturation	—	0.7494646681

Values with a unit may be written in any unit of the right kind.

Applies when

PO2 > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

Severinghaus saturation: PO2 = 40 mmHg

SO2 = 0.7494646681

Run it

Terminal

$ human-physiology solve hp_severinghaus_saturation "PO2=40 mmHg"
SO2 = 0.7494646681
  by: Saturation by the Severinghaus (1979) dissociation equation  (Severinghaus equation)
  PO2 = 5332.895497 Pa
  assumes: PO2 > 0

import human_physiology_ai999x as m

a = m.solve("hp_severinghaus_saturation", PO2="40 mmHg")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Serum anion gap and albumin-corrected anion gap1 fig.hp_anion_gap

AG = Na – (Cl + HCO3) and the albumin-corrected value AG + 2.5 (4.0 – albumin in g/dL).

AG = [Na^+] – ([Cl^-] + [HCO_3^-])

You give

	Meaning	Unit	Example
\text{Na}	Sodium (e.g. in mmol/L)	mol/m^3	140 mmol/L
\text{Cl}	Chloride (e.g. in mmol/L)	mol/m^3	104 mmol/L
\text{HCO3}	Bicarbonate (e.g. in mmol/L)	mol/m^3	24 mmol/L
\text{albumin}	Albumin in g/dL (optional; normal albumin if omitted) optional	—

You get

	Meaning	Unit	Example
\text{AG}	Anion gap in mmol/L	—	12
\text{AG\_corrected}	Albumin-corrected anion gap in mmol/L	—	12

Values with a unit may be written in any unit of the right kind. An optional input can be left out: a stated default is used.

Applies when

all inputs > 0

Outside these conditions it says what is wrong instead of returning a number.

Worked example

anion gap: Na = 140 mmol/L, Cl = 104 mmol/L, HCO3 = 24 mmol/L

AG = 12

AG_corrected = 12

Run it

Terminal

$ human-physiology solve hp_anion_gap "Na=140 mmol/L" "Cl=104 mmol/L" "HCO3=24 mmol/L"
AG = 12
AG_corrected = 12
  by: Serum anion gap and albumin-corrected anion gap  (AG = Na - Cl - HCO3; + 2.5 per g/dL of albumin below 4.0)
  Na = 140 mol/m^3
  Cl = 104 mol/m^3
  HCO3 = 24 mol/m^3
  assumes: all inputs > 0

import human_physiology_ai999x as m

a = m.solve("hp_anion_gap", Na="140 mmol/L", Cl="104 mmol/L", HCO3="24 mmol/L")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Static and dynamic compliance and airway resistance1 fig.hp_respiratory_mechanics

Static compliance VT/(Pplat – PEEP), dynamic compliance VT/(Ppeak – PEEP), airway resistance (Ppeak – Pplat)/flow and driving pressure.

C_{st} = \frac{V_T}{P_{plat} – PEEP},\quad R_{aw} = \frac{P_{peak} – P_{plat}}{\dot V}

You give

	Meaning	Unit	Example
\text{VT}	Tidal volume (e.g. in mL)	m^3	500 mL
\text{Ppeak}	Peak airway pressure in cmH2O	—	25
\text{Pplat}	Plateau pressure in cmH2O	—	20
\text{PEEP}	End-expiratory pressure in cmH2O	—	5
\text{flow}	Constant inspiratory flow (e.g. in L/min)	m^3/s	60 L/min

You get

	Meaning	Unit	Example
C_{\mathrm{static}}	Static compliance in mL/cmH2O	—	33.33333333
C_{\mathrm{dynamic}}	Dynamic compliance in mL/cmH2O	—	25
R_{\mathrm{aw}}	Airway resistance in cmH2O s/L	—	5
\text{driving\_pressure}	Driving pressure in cmH2O	—	15

Values with a unit may be written in any unit of the right kind.

Applies when

Ppeak >= Pplat > PEEP >= 0
constant flow

Outside these conditions it says what is wrong instead of returning a number.

Worked example

respiratory mechanics: VT = 500 mL, Ppeak = 25, Pplat = 20, PEEP = 5, flow = 60 L/min

C_static = 33.33333333

C_dynamic = 25

R_aw = 5

driving_pressure = 15

Run it

Terminal

$ human-physiology solve hp_respiratory_mechanics "VT=500 mL" Ppeak=25 Pplat=20 PEEP=5 "flow=60 L/min"
C_static = 33.33333333
C_dynamic = 25
R_aw = 5
driving_pressure = 15
  by: Static and dynamic compliance and airway resistance  (pressure differences over volume and flow)
  VT = 0.0005 m^3
  Ppeak = 25
  Pplat = 20
  PEEP = 5
  flow = 0.001 m^3/s
  assumes: Ppeak >= Pplat > PEEP >= 0; constant flow

import human_physiology_ai999x as m

a = m.solve("hp_respiratory_mechanics", VT="500 mL", Ppeak="25", Pplat="20", PEEP="5", flow="60 L/min")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Systemic oxygen delivery1 fig.hp_oxygen_delivery

DO2 = CO x CaO2 x 10 (mL O2/min), with CaO2 = 1.34 Hb SaO2 + 0.003 PaO2.

D_{O_2} = CO \times C_aO_2 \times 10

You give

	Meaning	Unit	Example
\text{CO}	Cardiac output (e.g. in L/min)	m^3/s	5 L/min
\text{Hb}	Hemoglobin in g/dL	—	15
\text{SaO2}	Arterial saturation	—	0.98
\text{PaO2}	Arterial PO2 (e.g. in mmHg)	Pa	100 mmHg

You get

	Meaning	Unit	Example
\text{DO2}	Oxygen delivery in mL O2/min	—	999.9
\text{CaO2}	Arterial oxygen content in mL/dL	—	19.998

Values with a unit may be written in any unit of the right kind.

Applies when

CO, Hb > 0
0 <= SaO2 <= 1

Outside these conditions it says what is wrong instead of returning a number.

Worked example

oxygen delivery: CO = 5 L/min, Hb = 15, SaO2 = 0.98, PaO2 = 100 mmHg

DO2 = 999.9

CaO2 = 19.998

Run it

Terminal

$ human-physiology solve hp_oxygen_delivery "CO=5 L/min" Hb=15 SaO2=0.98 "PaO2=100 mmHg"
DO2 = 999.9
CaO2 = 19.998
  by: Systemic oxygen delivery  (DO2 = 10 CO CaO2)
  CO = 8.333333333e-05 m^3/s
  Hb = 15
  SaO2 = 0.98
  PaO2 = 13332.23874 Pa
  assumes: CO, Hb > 0; 0 <= SaO2 <= 1

import human_physiology_ai999x as m

a = m.solve("hp_oxygen_delivery", CO="5 L/min", Hb="15", SaO2="0.98", PaO2="100 mmHg")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Temperature, pH and PCO2 correction of the dissociation curve (Severinghaus)1 fig.hp_virtual_po2

Virtual PO2 = PO2 x 10^(0.024 (37 – T) + 0.40 (pH – 7.40) + 0.06 log10(40/PCO2)) and the saturation from the standard curve (Bohr effect).

P_{virt} = P\cdot 10^{0.024(37-T) + 0.40(pH – 7.40) + 0.06\log_{10}(40/P_{CO_2})}

You give

	Meaning	Unit	Example
\text{PO2}	Measured PO2 (e.g. in mmHg)	Pa	80 mmHg
T	Temperature (e.g. in degC)	K	39 degC
\text{pH}	Blood pH	—	7.3
\text{PCO2}	PCO2 (e.g. in mmHg)	Pa	50 mmHg

You get

	Meaning	Unit	Example
\text{PO2\_virtual}	Standard-condition (virtual) PO2 in mmHg	—	64.45778629
\text{SO2}	Fractional saturation	—	0.9222276859
\text{factor}	Correction factor	—	0.8057223286

Values with a unit may be written in any unit of the right kind.

Applies when

25 <= T <= 43 degC
6.8 <= pH <= 7.8

Outside these conditions it says what is wrong instead of returning a number.

Worked example

virtual PO2: PO2 = 80 mmHg, T = 39 degC, pH = 7.3, PCO2 = 50 mmHg

PO2_virtual = 64.45778629

SO2 = 0.9222276859

factor = 0.8057223286

Run it

Terminal

$ human-physiology solve hp_virtual_po2 "PO2=80 mmHg" "T=39 degC" pH=7.3 "PCO2=50 mmHg"
PO2_virtual = 64.45778629
SO2 = 0.9222276859
factor = 0.8057223286
  by: Temperature, pH and PCO2 correction of the dissociation curve (Severinghaus)  (Severinghaus (1979) correction, then the standard curve)
  PO2 = 10665.79099 Pa
  T = 312.15 K
  pH = 7.3
  PCO2 = 6666.119371 Pa
  assumes: 25 <= T <= 43 degC; 6.8 <= pH <= 7.8

import human_physiology_ai999x as m

a = m.solve("hp_virtual_po2", PO2="80 mmHg", T="39 degC", pH="7.3", PCO2="50 mmHg")
print(a.values or (a.value, a.unit))

Figure

Each figure varies one input around the worked example and recomputes the result at every point. Press Compute above and they are redrawn for your own inputs.

Figures

Computed, not drawn by hand

Every point in these figures is an answer of the model: one input is varied, the others are held at the worked example, and the curve shows how the result responds. The marked point is the worked example itself.

Nernst equilibrium potential of an ion

E = \frac{RT}{zF}\ln\frac{[C]_o}{[C]_i}

E (equilibrium potential in mV) against Cout (extracellular concentration in mM), with Cin = 140, z = 1. 31 points computed by human-physiology-ai999x; marked: Cout = 4, E = -95.0226.

Body fat from body density (Siri and Brozek)

\%BF = \frac{495}{D} – 450

fat_Siri (body fat in % (Siri)) against density (body density (e.g. in g/mL)). 9 points computed by human-physiology-ai999x; marked: density = 1050, fat_Siri = 21.4286.

Two-compartment fluid shift after adding a solution (Darrow-Yannet)

P_{osm}' = \frac{TBW\,P_{osm} + V C}{TBW + V},\quad ICF' = \frac{ICF\,P_{osm}}{P_{osm}'}

osm_final (final osmolality in mOsm/kg) against TBW (initial total body water (e.g. in L)), with C_added = 0, Posm = 290, V_added = 0.002 m^3. 31 points computed by human-physiology-ai999x; marked: TBW = 0.042, osm_final = 276.818.

Bernoulli pressure drop across a narrowing

\Delta P = \tfrac{1}{2}\rho\,(v_2^2 – v_1^2) \approx 4 v_2^2

dP_mmHg (pressure drop in mmHg) against v1 (upstream velocity (e.g. in m/s)), with rho = 1060 kg/m^3, v2 = 4 m/s. 31 points computed by human-physiology-ai999x; marked: v1 = 1, dP_mmHg = 59.6299.

Versions & files

Releases and repository files

The source is on GitHub, with every release: github.com/AI999X/human-physiology-ai999x ↗

Files

19 files, 41.8 MB in total. The largest:

File	Size
model.safetensors	18.9 MB
dist/human_physiology_ai999x-1.0.0-py3-none-any.whl	7.2 MB
onnx/search/index.json	6.5 MB
onnx/search/encoder.int8.onnx	4.9 MB
docs/human_physiology_ai999x_capabilities.pdf	3.6 MB
docs/human_physiology_ai999x_capabilities.html	298.8 KB
docs/human_physiology_ai999x_capabilities.md	224 KB
onnx/search/tokenizer.json	173.3 KB
README.md	36.8 KB
LICENSE	11.5 KB
figures/hp-bernoulli-pressure-drop-dp-mmhg-vs-v1.svg	3 KB
figures/hp-nernst-potential-e-vs-cout.svg	2.9 KB

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Questions

Frequently asked

What does human-physiology-ai999x do?

Human Physiology answers questions in human physiology: cardiac output, blood pressure, vascular resistance and Poiseuille flow, lung volumes, ventilation and the alveolar gas equation, oxygen content and binding, acid-base, renal clearance and eGFR, Nernst and Goldman-Hodgkin-Katz potentials, cable properties, body composition and metabolic rate (computed values, not medical advice). Each answer gives the value with its unit, the steps and the formula in LaTeX.

Is there a desktop app or a command-line program?

Yes, both, for Windows and Linux, in their next release: the AI999X desktop app holds all AI999X models in one window, and the command-line bundle one program per model (human-physiology for this one). Until then this model runs in your browser on this page. Downloads and checksums: https://www.ai999x.com/desktop-cli/

How do I install human-physiology-ai999x?

With pip: pip install human-physiology-ai999x. The same files are on Hugging Face (hf download AI999X/human-physiology-ai999x –local-dir human-physiology-ai999x) and GitHub.

Does it run offline?

Yes. It is a Python package that runs on an ordinary CPU without a network connection (Python >=3.10); no GPU and no API key are needed.

How many capabilities does human-physiology-ai999x have?

96 capabilities in 7 topics, including Respiratory, Cardiovascular, Nerve and muscle, Renal, Metabolism.

Which field and tasks does it cover?

Field: Life Sciences. Tasks: Question Answering, Sentence Similarity, Feature Extraction.

What does an answer look like?

Asked "cardiac output: HR = 72, SV = 70 mL", it answers: CO = 5.04, CO_mL = 5040.

What happens when a question cannot be answered?

It says so and names what is missing or out of range, instead of guessing a number.

Explore every model.

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