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winter camping, air-matress-tent inside of a tent, & a candle

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Posted by zoe_lithoi on October 10, 2016, 12:06 am
I think I've finally found a way to build a tent that will keep you plenty  
warm in the winter, and yet be small enough (once it is folded up) to easil
y transport in a small part of the trunk of your car.

The idea is to make a tent out of air-mattresses which is placed inside of  
a normal tent. alternatively, one might buy and make use of an inflatible t
ent such as the 'Cave' from Hemiplanet ($00) :  www.heimplanet.com

Caution though, I am not sure if just the poles are inflatible, or the whol
e thing. Further, if this were to be used, it would still need to be placed
 inside of another tent, and the calculations adjusted, from those within t
his posting to account for the dimensions.....

 The air-mattress tent inside of another tent can keep you plenty warm  and
 this can be done without a gas heater and without candles ---- though with
 the possibility of condensation and if there are no 'leaks', then carbon m
onoxide might be an issue. To avoid the latter 2 problems, airflow vents an
d a candle would have to be added.

 I have a quick word of praise about candles. In the past, I looked at usin
g tea-candles for heating a well-insulated room with 4 or so people in it.  
The tea-candles cost about $.65 for 100 and  produce about 250 Btu/hr and  
last for 4 hour's. Nevertheless, there are some disadvantages. Candles prod
uce smoke, soot, and carbon dioxide. these issues can be elliminated with v
ents for airflow....  

as many of you experienced campers know, that sometimes a tent will get wet
 on the inside due to condensation. The condensation occurs when the vapor  
in the air forms water on the inside tent surface. The water in the air pri
marily comes from you breathing it out into the air. If you have a cold sur
face and warm air, you can get condensation. This can actually be determine
d using such methods as 'dew point' and wet and dry bulb temperatures. But  
that is outside the scope of this posting. The concept is, however, that if
 you keep the surface warm enough, or the air cold enough, or reduce the va
por in the air low enough, the condensation won't occur.

So, you might stop possible condensation by having a candle provide heat fo
r the air space between the two tents (the regular tent and the air-matress

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 one from the ground.

Since the winter time ground temperature is warmer than the air temperature
, I think the bigger question should be, can an aire mattress be used to in
sulate 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
oes not.

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:

Tbdy(100F)---www (Rsa-air-mat)---www(Rfab-air-mat)---||(Cair)---Tgnd(50degF
---- 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
--- RC-Time-Constant
= 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
-still-air' resistances.
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
(100-Ttent)*3 = (Ttent - 30)*9
300 - 3Ttent = 9Ttent - 270
570 = 12Ttent
Ttent = 570/12 = 47.5F.

Now, the air outside your sleeping bag is 50degF instead of 30degF


What happens when we put this air mattress tent inside of another tent (of  
temp, Tout-temp)?

The outer tent thermal resistance:
a 5x7x6tall 4sided tent has a surface area of:
4x5x7 + 6x7 = ~200sqft

Each surface (inner and outer has a warm-still-air resistance.
Router-tent = 2 x (2/3sqft-Hr-F/Btu ) / 200 sqft = 0.0067

Let's assume our body temp is 100degF

Heat(body) = (Tbdy - Tgnd)/Rbag-mat + (Tbdy - Ttent)/Rbag-tent + (Ttent-3
300Btu/hr = (100 - 50F)/0.5 + (100 - Ttent)/0.44 + (Ttent-30)/0.007
300 = (50)*2 + (100 - Ttent)*2.3 + (Ttent-30)*150
300 = 100 + 230 - 2.3Ttent + 150Ttent - 4500
4470 = 148Tout-tent
Tout-tent = 4470 /148 = 30.2degF

So, it's just a little bit warmer than the outside temperature.


In the real world, there are air leaks between tent openings etc. Some tent
 openings are there on purpose to prevent condensation. as the air flows, i
t takes out heat.  You need a 6 air-exchange per hour rate for healthy brea
thing anyways. The airflow could be accomplished if the inside of our air-m
attress tent has an inlet opening to the outside, and the top corner of the
 air mattress tent has an opening to the air space to the normal tent.... t
he normal tent already has an opening at the top.

The volume of the tent might be 5 x 6 x 6 = 180cuft
For airflow, 6 times this is:  ~1000 cuft/hr

The amount of heat being lost to air exchange is:
1000cuft/hr x (Ttent - Tair)degF x 1/55 Btu/cuft/degF
1000 x (50 - 30) x 1/55 Btu/hr = ~400Btu/hr
Heat(air-exchange) = 400Btu/hr

Since the body is only producing 300 btu/hr, then one can see a big problem
. Your body will lose heat. Hence, one would need use a candle to provide e
xtra heat.

If we were to put a 250btu/hr brittish tea-candle in between the air space  
between the outer mattress surface, and the inside surface of the regular t
ent, then it would help with the heating situation as well as possible cond
ensation issues.

Now, factoring this into the circuit

Heat(body) + Heat(Candle) - Heat(air-exchange) = (Tbdy - Tgnd)/Rbag-mat +
 (Tbdy - Ttent)/Rbag-tent + (Ttent-30)/Router-tent
300Btu/hr + 250Btu/hr - 400btu/hr = (100 - 50F)/0.5 + (100 - Ttent)/0.44  
+ (Ttent-30)/0.007
150 = (50)*2 + (100 - Ttent)*2.3 + (Ttent-30)*150
150 = 100 + 230 - 2.3Ttent + 150Ttent - 4500
4680 = 148Tout-tent
Tout-tent = 4680 /148 = 31.6 degF

It is only 1.6degF warmer than outside

That means the air temp between the inner tent (air-mattress) and outer ten
t is warmer than the 30degF outside air.... and we now have enough airflow  
to hopefully prevent condensation and also to provide protection against ca
rbon monoxide poisoning.

What is the temperature of the air inside the inner tent (air-mattress)?

(Tbdy - Ttent)/Rbag = (Ttent - Tout-tent)/Rtent-air-mat
(100-Ttent)/0.333 = (Ttent - 31.6)/0.11
(100-Ttent)*3 = (Ttent - 31.6)*9
300 - 3Ttent = 9Ttent - 284
584 = 12Ttent
Ttent = 597/12 = 48.7F.

It is only 1.2degF warmer than the 47.5degF we got without considering the  
outer tent, the candle, and the air-exchange.


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