I've been venturing into hunting the last couple years, in the great southw
est part of america. Some hunts are in November to January. So it does get
cold --- not Minnesota cold, but cold none-the-less ---- say 30degF.
Is it a good idea to use an air mattress as ground insulation to keep 'warm
and if so, could an air mattress 'tent' be used to keep warm when camping?
Why an air mattress?
----- it can be packed up into a small box and easily (relatively) transpo
rted in your car on your camping/hunting/fishing trip.
a cursory look at the internet shows various opinions (none with scientific
temperature data or even any calculations ---- some articles think it cri
minal to suggest that an air mattress even be suggested as a means to insul
ate from the ground.
Since the winter time ground temperature is warmer than the air temperature
, I think the bigger question should be, can an air mattress be used to ins
ulate one from the cold air?
.... or perhaps even the biggest question, can an air mattress be used as i
nsulation at all?
Let's try some calculations.
The amount of heat which air can hold
---- thermal air capacitance ---- is:
Cair = 1/55 Btu/cuft/degF
The volume of air inside a 3ft x 6ft x 0.25ft air mattress is:
Vair-mat = 3x6x0.25= 4.5cuft
Cair x Vair-mat = 1/55 Btu/cuft/degF x 4.5cuft = ~ 1/15 Btu/degF
So, even if the temperature was 90degF, it would only hold 90/15=6 Btu's
Not very much. ....air is not a good capacitor. It doesn't store much heat
--- unless you have alot of air flow, which in this case, an air mattress d
Nevertheless, the air matress provides a top and bottom surface which can b
e greatly used to provide resistance layers of what is called warm-still-ai
r-resistances. ...and these we will see, help tremendously.
When one thinks of 'insulation', one thinks of 'R-Value', which has a close
relative called 'thermal resistance'.... which simply means how slow or fa
st something resists heat flowing through it. What units of measurement? We
ll. 'slow or fast' indicates 'time', ie minutes or hours. 'flow' and 'heat
flow' indicate amount of heat (BTU) per degree F, so the units would be:
Hr x degF/Btu (often written in a shorter notation as: Hr-degF/hr)
In addition, thermal resistance is defined as the amount of heat flowing th
rough a certain cross-sectional area, (which in this case is your 1.5ft x 6
ft body = 9sqft). So the units would be:
Looking at the air in the inside of the air mattress, there are 2 surfaces
- 1 on top, and one on the bottom.
In general, the thin layer of still air thermal resistance on a surface, Rs
a, has been measured at about
Rsa = the still air resistance = 2/3 sqft-Hr-F/Btu
So, the 2 air mattress surfaces have a thermal still air resistance of abou
Rsa-air-mat = 2 x Rsa x Area = 2 x (2/3sqft-Hr-F/Btu ) / 9sqft
= 0.15 Hr-F/Btu
The fabric of the air matress also has thermal resistance
total thermal resistance,
= the air-resistance of these 2 surfaces (Rsa-air-mat)
+ the resistance of the fabric of the 2 surfaces (Rfab-air-mat)
+ the air-resistance of the 3inches of air inside the air mattress (I've re
ad that this is negligible)
Clothing has a thermal resistance of:
Rcloth=3.0xth sqft-F-hr/Btu; where: th is in units of 0.1 lb/sqin.
I'm guessing the air mattress might be made out of something like 'balloon
cloth' which weighs very roughly 4oz/sqyd x 1 lb/16oz x (1yd/36in)^2 = 0.
0002 lb/sqin = 0.002 x 0.1 lb/sqin
Rcloth = 3 x 0.002 = 0.006 sqft-F-hr/Btu
For our air mattress, there are 2 layers of cloth each 3'x6'(18sqft), hence
Rfab-air-mat = 2 x 0.006 sqft-F-hr/Btu /18sqft = 0.0007 F-hr/Btu
Remembering, from above, the thermal capacitance of the air, and taking int
o account the body temperature of the human body laying on top of the mattr
ess, called Tbdy, and for now ignoring the insulation of a sleeping bag,
is roughly 100degF, and estimating the ground temperature might be very ro
ughly, Tgnd = 50degF
The reason we are 'ignoring' the sleeping bag insulation for now, is becaus
e we are just trying to characterize the air mattress.
Letting the symbol '-www-' represent resistance, and '-||-' be capacitance,
a circuit model might be something like:
---- Cair=1/10 Btu/degF
---- Rsa-air-mat= 0.15 Hr-F/Btu
---- Rfab-air-mat = 0.0007 F-hr/Btu
So, it's fairly obvious that the resistance of the fabric is way smaller th
an that of the still-air.
The above is the classic RC circuit:
Tair-mat(t) = Tair-mat(t=0) + (Tair-mat(max)-Tair-mat(t=0) )(1-e^-t/R
= Rsa-air-mat=0.15 Hr-F/Btu x Cair=1/10 Btu/degF = 0.015 hr
This is so small, that the thermal capacitance is really negligible.
The air is between the 2 surfaces of the air mattress, so in reality, the c
ircuit looks like:
Tbdy(100F)--www (Rsa-air-mat/2)--Tmat-www (Rsa-air-mat/2)--Tgnd(50degF).
It should be obvious that Tmat will be 1/2 way between 100 and 50, hence T
mat = 75degF
The heat flowing through the mattress is:
(Tbdy-Tgnd)/Rsa-air-mat = (100-50)F/0.15 Hr-F/Btu = 333Btu/hr
Your body, at rest, generates about 300BTU/hr of energy (heat)
Most of that will go into the air, and there won't be enough to provide 333
BTU/hr, so your body temperature will get colder.....
One must use a sleeping bag, which has an R-Value of about 3 sqft-F-hr/Btu.
Half of that is on the top of your 9sqft body, and half on the bottom.
Rbag = 3 sqft-F-hr/Btu / 9sqft = 0.333 F-hr/Btu
Now you can appreciate that the air matress's 0.15 hr-F/Btu thermal resista
nce is 50% of a sleeping bag. I feel that is significant. The combined slee
ping bag and air matress resistance (between your body and the ground is:
Rbag-mat = 0.15 + 0.33 = 0.48 = approix. 0.5 hr-F/Btu.
the thermal resistance between your body and the air, is just the sleeping
bag, ie. Rbag
The circuit might look like:
Heat(body) = (Tbdy - Tgnd)/Rbag-mat + (Tbdy - Tair)/Rbag
300Btu/hr = (Tbdy - 50F)/0.5 hr-F/Btu + (Tbdy - 30F)/0.333hr-F/Btu
300 = (Tbdy - 50)*2 + (Tbdy - 30)*3
300 = 2Tbdy - 100 + 3 Tbdy - 60
460 = 5Tbdy
Tbdy = 460/5 = 92degF
Your body will become colder than it's 98.6degF.
If we didn't use the air-mattress, it would be:
300Btu/hr = (Tbdy - 50F)/0.333 hr-F/Btu + (Tbdy - 30F)/0.333hr-F/Btu
300 = (Tbdy - 50)*3 + (Tbdy - 30)*3
300 = 3Tbdy - 150 + 3 Tbdy - 60
510 = 5Tbdy
Tbdy = 510/6 = 85degF
Without an air mattress, your body gets even COLDER! (From 92 to 85F).
Now what happens, if we make an air-mattress tent around us.
The air-mattress 'roof' now even works better because now there are 5 'warm
1. between the outside air and upper fabic
2. on the inside of the upper fabric
3. on the inside of the lower fabric
4. between the outside of the upper fabric and the air in your tent
5. between air in your tent and the sleeping bag.
So the thermal resistance of the 'roof' air mattress is:
Rroof-air-mat = 5 x Rsa x Area of mat
The area of the mat includes the 3x6 roof as well as both 2x3 sides, hence:
3x6 + 2x2x3 = 18 + 12 = 30sqft
Rtent-air-mat= 5 x (2/3sqft-Hr-F/Btu ) / 30 sqft
= ~ 0.11 Hr-F/Btu
The total resistance between your body and the outside is now:
Rbag-tent = Rbag + Rtent-air-mattress
Rbag-tent = 0.33 hr-F/Btu + 0.11 hr-F/Btu = 0.44 hr-F/Btu
The circuit is now:
Heat(body) = (Tbdy - Tgnd)/Rbag-mat + (Tbdy - Tair)/Rbag-tent
300Btu/hr = (Tbdy - 50F)/0.5 hr-F/Btu + (Tbdy - 30F)/0.44hr-F/Btu
300 = (Tbdy - 50)*2 + (Tbdy - 30)*2.3
300 = 2Tbdy - 100 + 2.3Tbdy - 69
469 = 4.3Tbdy
Tbdy = 460/5 = 109degF
Your body will be really warm, and you will probably open up your sleeping
Looks look at the temperature of the air inside the tent, Ttent
(Tbdy - Ttent)/Rbag = (Ttent - Tair)/Rtent-air-mat
(100-Ttent)/0.333 = (Ttent - 30)/0.11
(109-Ttent)*3 = (Ttent - 30)*9
327 - 3Ttent = 9Ttent - 270
597 = 12Ttent
Ttent = 597/12 = 50F.
Now, the air outside your sleeping bag is 50degF instead of 30degF