# Is the skydiver cold?

1. Jan 30, 2005

### Mk

I've never gone skydiving, but, if you jump off an airplane, will it be hot or cold, or neither? I was thinking, ram pressure and friction from the air would heat you up, though wind cools you off by carrying away heat... hot or cold?

2. Jan 30, 2005

### buddingscientist

id go colder. less atmosphere so less heat being radiated back into the earth

on another note, i recall on an airplane trip they had a screen with velocity, temp etc and the outside temperature was -52deg C (at peak/near peak commercial airliner altitude)

3. Jan 30, 2005

### Staff: Mentor

There is a temperature gradient - the higher you go, the colder it is. On a nice warm summer day, its in the 40s at the altitude you jump from.

There is nowhere near enough wind resistance to create noticeable frictional heating.

4. Jan 30, 2005

### DaveC426913

Colder. Ram pressure and friction are negligible compared to the temperature at altitude and the heat loss due to wind.

temp vs. altitude graph

The temperature drops dramatically as altitude increases. I don't know what standard height for skdiving is, but it's below 30,000 feet for sure.

5. Jan 30, 2005

### Staff: Mentor

12,000-14,000 feet, Dave.

6. Jan 30, 2005

### DB

Why are the stratosphere and thermosphere relatively hot?

7. Jan 31, 2005

### Akihiro

Maybe because of bombardment from cosmic rays and UV rays or because the gas are more free to move about at altitudes where gravity starts weakening... I'm not quite sure though.

8. Jan 31, 2005

### Clausius2

The stratosphere is hot due to photochemical reactions. In particular, the presence of the Ozone layer absorbs the Ultra Violet radiation of shortest wave lenghts.

Reactions of Ozone destruction such us ozone photolisis:

$$O_3 + h\nu \rightarrow O_2 + O + Energy$$

and molecular oxygen destruction (molecular oxygen photolisis):

$$O_2 + h\nu \rightarrow O + O + Energy$$
$$O_2+O+M\rightarrow O_3 +M$$ where M is some reaction activator,

Both of them causes a heating of the stratosphere and a thermal inversion.

As far as the Thermosphere is concerned, I have seen large variations in its temperature. At heights of 400 km or so, where Space shuttle and satellites usually orbit, there are 1200 ºC or so. But that temperature is not able to heat up spatial structures, due to the poor concentration of molecules there. The kinetic energy transference of such molecules is very small, and so that the heat produced.

9. Jan 31, 2005

### Clausius2

Maybe doing some number of simple approximation we could fade away your doubts. Let's suppose the skydiver as cylinder shaped of diameter D=30 cm and lenght L=1.7 m.

-Thermal power absorbed due to friction: according to usual estimations, the friction coefficient is about $$C\approx 0.3$$. The thermal power will be approximately:

$$\dot Q_f \approx 1/2 \rho_{air} \cdot DL \cdot C \cdot U^3$$

where U is an average speed, let's say: $$U\approx 100 km/h$$.

- Thermal power lost due to convection. The external temperature will be at $$T_{\infty}=-50 ºC$$ as the graph of DaveC426913 suggests at 10 km of altitude.

The Reynolds number of the flow will be:

$$Re=\frac{UD}{\nu}=\frac{100\cdot 0.3}{35.2\cdot 10^{-6}}=852272$$ which is fully turbulent. The thermal properties of the air have been obtained from the Standard Atmosphere.

The Prandtl number of the flow will be:

$$Pr=\frac{\nu}{\alpha}\approx 0.7$$

According to Incropera ("Heat Transfer Fundamentals" Ed. Prentice Hall), experimental results show that the Nusselt number can be obtained from the next correlation:

$$Nu_D=\frac{hD}{k}=0.027Re^{0.805} Pr^{1/3}=1425$$ where h is the convection coefficient and k is the thermal conductivity of the air. So that, we obtain:

$$h\approx 125 \frac{W} {m^2ºC}$$

Therefore the thermal power lost by convective effects is:

$$\dot Q_{c}=\pi D L h (\overline{T}-T_{\infty})$$

where $$\overline{T}$$ is an average temperature of the body, let's say 20ºC. So that, let's compare both powers:

$$\frac{O(friction)}{O(convection)}=\frac{1/2 \rho_{air} \cdot C \cdot U^3}{\pi h 80 }\approx 0.11$$

What is the conclusion?. After spending the effort, I have realised that's a waste of time :rofl: . The fact is that high freefalling velocities such Space Shuttle reentries will enhance a greater heating due to friction. But small velocities like that experimented by an skydiver surely won't heat him up.

I'm not sure. What do you think? :shy:

10. Feb 1, 2005

### DB

Thank Clausius, though wats "hv" stand for, high frequency ultra-violet rays??

11. Feb 2, 2005

### Clausius2

h=Planck's constant.
v=frecuency of the ray. (I think it's 10e16 Hz for UVB radiation or so).

12. Feb 2, 2005

### krab

He's cold alright. You might want to look at this incident:

10 June 1990: A British Airways British Aircraft Corporation (BAC) 111 makes an emergency landing at Southampton after its front windscreen blows out and its captain is partially sucked out of the flight deck during depressurisation.

The captain had his whole body sucked out, and a steward was hanging on to his legs while the co-pilot made an emergency landing. He was assumed by everyone to be dead, but he survived with no serious injuries. I guess he was frozen quickly enough that this lowered his body's need for oxygen.

13. Feb 4, 2005

### DB

Thanks, sry for the delay I was away for a couple of days. Now Im not saying you're wrong cuz Im sure you're right. But I've read that the ozone is what protects us from the sun's UV rays, but with the equation:
$$O_3 + h\nu \rightarrow O_2 + O + Energy$$
It seems that any ray can split O3. Im just a little confused here.

14. Feb 5, 2005

### Clausius2

Yes, you're right. The Ozone absorbs the UVC and UVB radiation, by means of the Ozone destruction reaction I have posted. The O3 absorbs the radiation hv (which merely means UVC and UVB radiation, it is formalism), splitting into molecular oxygen and atomic oxygen. What is what you don't understand?

15. Feb 5, 2005

### DB

1. Where does the UV radiation go? Is it refracted?

2. Can ozone re-form after being split?

16. Feb 5, 2005

### Grogs

In the US, we use the approximation that every 1000 feet of altitude decreases the temperature by 2.5 degrees. This is of course an approximation based on an idealized atmosphere, but it is accurate more times than not. According to the approximation I mentioned above, a 15,000 foot difference in elevation (assuming your dropzone was near sea level) would result in a 35-40 degree temperature drop. Because of FAA (and I'm pretty sure the FAI has similar requirements) to have oxygen above 15,000 feet, almost all recreational skydives happen at or below 15,000 feet MSL.

I can speak to this from first hand experience. On a hot 90 degree day in the middle of Summer, the outside air feels very pleasant at 15,000 feet. Think of it like walking into a giant refrigerator on a Summer's day. You'd get cold after several minutes there, but it feels really good at first. A lot of skydivers typically jump in shorts/t-shirts in the summer. Typical freefall time for a jump like this would be ~60 seconds, btw.

As temperatures start dropping though, the conditions at altitude become more and more unpleasant. Heat from friction is really not a factor (that I could ever tell.) Rather, I think of it more like a 120 mph wind chill. On a pleasant fall day (50's) I have to wear gloves to prevent my hands from getting numb. In the cold days of winter, you just have to bundle up. Any exposed skin will get frostbitten.

17. Feb 5, 2005

### Clausius2

1.I'm not an expert on this stuff, but I think the UV radiation will be absorbed by the Ozone molecule, and thus this molecule is destructed.

2. Yes, the ozone is reformed with the molecular oxygen photolisis reaction.
The problem is that the warming effect gases (CFC, CO2, CH4) destroy the ozone much faster.

18. Feb 5, 2005

### DB

Thanks, do you know where, once absorbed, the energy of the UV wave is then sent? I think it must be conserved, but once the molecule is split, where is the release in energy sent?

19. Feb 5, 2005

### Clausius2

The wave energy is not "conserved". The ozone molecule breaks up releasing some energy. The fact is that the ozone destruction reaction is exothermal, so that it is one of the direct responsible of the thermal inversion (heating) in the Stratosphere.

20. Feb 5, 2005

### DB

Ahhh, I see it now. I've researched the oxygen ozone cycle as you have showed me.
As UV radiation from the sun hits earth, it splits O3 molecules, releasing energy (heat). Then O2 may split or react with a lone O atom, forming the 3 electron pairs, then being split again, so on and so forth, heating up the stratosphere. Thanks.