# Ruben was right

```Here I with numbers in hand try to confirm John Ruben's claim about
respiratory water loss being a real danger to endothermic vertebrate
without special means of preventing water loss. The only thing
found is that it only applies if lung temperature is above 43°C.
Ruben is entirely correct in sense that it is highly improbable that
dinosaur body temperature was above 43°C.```

```Table: Pressure of saturated water vapour
(some points are omitted)
T,°C kPa
20 2.3368
28 3.7785
30 4.2417
32 4.7536
34 5.3182
36 5.9401
38 6.6240
40 7.3749
42 8.1938
44 9.0998
52 13.612
54 15.001```

```There are three sources of water:
- water that animal drinks
- water present in food
- oxygen metabolism```

```Some food is not dissasimilated. It supplies organism growth.
As it is not oxygenated, it does not influence respiration nor
provides extra water, and therefore is no subject of this analysis.
If animal produces mixed fecal-urine mass which has same
water fraction as food, that loss of water can be also neglected.
We can equate it as food that has not been eaten.```

```I assume that partial pressure of carbon dioxide in exhausted
air is 4%. We can't be sure which value extinct animals did have.
Atmospheric pressure is about 101 kPa (after all, most animals
live near sea level). Various diving animals usually have much
larger values, so it shows it is not unusual to develop that.```

```First case.
Supoose we have a generic animal without RT...
Let's assume our generic animal eats 40% glucose + 60% water food
and does not drink any water, ambient air it totally dry and exhausted
air has moisture of 100% and 4% CO2.```

`101 kPa * 4% = 4.04 kPa. pressure of CO2`

For some period of time our generic animal produces enough CO2 to make its pressure in
its lungs 4.04 kPa and is now ready to exhale it.
Remember that glucose brutto-formula is C6H12O6.
Recall ideal gas equation. So if animal somehow produced
CO2 in amount to make 4.04 kPa pressure in a given volume (lungs), it
has to produce H2O in amount that would make 4.04 kPa pressure
in same volume. Note that in case of fat or protein food there is
slightly more metabolic water.

```Look at the table of saturated water vapour pressure...
Near T=29°C there should be equilibrium between metabolic water
and evaporation in lungs. Below T=29°C, metabolic water is more
than respiratory water loss. Monotremes have normal body T like 25--30°C.
They are (or considered to be) endotherms.```

```Now, consider that our animal has also water that was in liquid form
and is now free because glucose has been oxygenated. In C6H12O6,
"water" part consists
(12 + 6*16)/(6*12 + 12 + 6*16) = 3/5 of mass, or it is
3/5 * 40% = 24% of food mass.
In food where was 60% of liquid water. There is
60% / 24% = 2.5
times more free water than metabolic water. If free and metabolic
water is all lost though respiration... The animal could
afford to have H2O pressure of
4.04 + 4.04*2.5 = 14.14 kPa
Look at the table. We are somewhere near 53°C...```

```On that diet, our generic animal gains more water that it loses
if its lungs T is below 53°C.```

```Second case.
Let's assume our generic animal eats 40% fat + 60% water food.```

```We might assume that fat has brutto-formula (CH2)_n. This makes
us to overestimate respiratory water loss and underestimate
metabolic water, but we pass on for simplicity.
The metabolic water has "pressure" of 4.04 kPa as with glucose.
Let's count amount of free water.
In CH2, "H" consists
2/(2+12) = 1/7 of mass and
metabolic water is
1/7*(16+2)/2 = 9/7 of fat mass.
That means free water is
60%/40% * 7/9 = 7/6 times more than metabolic water.
The animal can therefore afford to have H2O pressure in his lungs...
4.04 + 7/6*4.04 = 8.753 kPa
Look at the table. We are somewhere near 43°C...```

```On that diet, our generic animal gains more water that it loses
if its lungs T is below 43°C.```

```Aminoacids are a bit harder to estimate...
I think result is beetwen that of carbohydrates and fat,
closer to that of fat.```

```It should be noted that above is worst-case except air pressure.
But if we change air pressure then we can't simply used value 4%.
In reality, the lung T is lower than internal organs T.
There is some moisture in ambient air.
Some animals have more than 4% CO2 in exhausted air at sea level.```

```Metabolism speed is not important. The important factor
is partial pressure of carbon dioxide in exhausted air.
If an ectotherm somehow managed to obtain high body T with any
kind of "gigantothermy" it would face difficulties with respiratory
water loss strictly equal to those in endotherms with same lung
temperature.```

```Herbivores need to dump excessive minerals present in plants,
and carnivores need to dump excessive nitrogen. But, we assumed
our generic animal did not drink. So it doesn't change much.```

```Effective methods of coping with water loss (if ever needed) are:
Raising CO2 pressure in exhausted air.
Lowering lung temperature.```