Interesting practical question skydiving

  • Thread starter zwoodrow
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So Ive read the highest skydive occured from a ballon and the entire jump lasted on the order of 10s of minutes. Heres my question: How the do you know where your gonna land? The earth rotates at ~1000mi per hour thats 12 mi per minute or 2.7 miles per 10 secs. So did these people use radar or something to time the exact time to jump? If you jump ~30 sec off target you land 10 miles off course. Does anyone know how they accounted for this? Say you have a bit of a problem jumping and have to wait 1.5 minutes to jump- your 30 miles off target.
 

rcgldr

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For the weather balloon jumps, the issue is prevailing winds at high altitudes, which can be up to 100 mph. In Kittenger's jump, the climb covered a lot of distance, but the winds in the drop mostly canceled so it wasn't an issue. The second link includes a map:

http://en.wikipedia.org/wiki/Joseph_Kittinger

http://stratocat.com.ar/fichas-e/1960/HMN-19600816.htm

For some what high jumps from aircraft, math similar to that used by bombers is used, so the skydivers don't have to compensate much for the initial drop. Once in the air, skydivers can maneuver quite somewhat to "glide" to a target, but the glide ratio is less than 1:1. With a wing suit, it's about 2.5:1, but it's not common to use wing suits for very high jumps
 

diazona

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The Earth's rotation has basically nothing to do with it. Remember that not only does the ground rotate, but the atmosphere rotates along with it as well. If you're up in a balloon, the balloon is also rotating along with the atmosphere (and the Earth). If you jump out of the balloon, you are already moving along with the balloon, and you keep moving along with the balloon as you fall to the ground. (well, almost; I'll refrain from explaining why it's not exact unless you're interested)

The bottom line is, when you jump out of a balloon, if there were no wind, you could expect to land pretty nearly at the point on the ground that is directly below the balloon. If there is some wind (which is generally the case in reality), then that will blow you off course, sure, but that doesn't depend on when you jump.
 
The guy that did this jump rode a baloon up 10+ miles in the atmospher where the air pressure was 1/80 He essentally had no atmosphere to speak of. Where does your extrapolation end? 10 miles 100 miles off the surface if your half way between the earth and the moon what your saying doesnt count so there is some point where the atmospheric drag can be neglected? I think the posts from the previous reply indicate that if you didnt have a peculular mix of countervaling winds in the troposphere and stratospher you would end up really far off course.
 

diazona

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If there was no atmosphere to speak of, how did the balloon stay up? :wink:

There is a point at which the atmosphere can be neglected, but it's higher than 10 miles. Depending on what exactly you are neglecting, I think it'd probably be something like 50 or 100 miles. Once you get too high for a balloon to support you, you have to be in orbit to keep from falling, and there is a particular speed (dependent on altitude) with which you have to move in order to keep from falling out of orbit. It still has nothing to do with the Earth's rotation, though.

It's possible to measure the wind speed at various altitudes and thus get a rough idea of how far they would blow the guy sideways. Yes, it is unlikely that the winds will exactly cancel out, but it's also unlikely that they will all add together in the same direction. The most likely case is that they'll partially cancel out, so he gets blown a moderate distance.
 
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The guy that did this jump rode a baloon up 10+ miles in the atmospher where the air pressure was 1/80 He essentally had no atmosphere to speak of. Where does your extrapolation end? 10 miles 100 miles off the surface if your half way between the earth and the moon what your saying doesnt count so there is some point where the atmospheric drag can be neglected? I think the posts from the previous reply indicate that if you didnt have a peculular mix of countervaling winds in the troposphere and stratospher you would end up really far off course.
Even if there was no atmosphere and the balloon magically stayed up, it still started out on the ground with the same 1000mph speed as the earth (at the equator).

So, why would it lose this speed on its way up? At the ground, near the ground, 10 miles up...it's still going to have the same sideways speed unless acted upon by an outside force.
 
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Even if there was no atmosphere and the balloon magically stayed up, it still started out on the ground with the same 1000mph speed as the earth (at the equator).

So, why would it lose this speed on its way up? At the ground, near the ground, 10 miles up...it's still going to have the same sideways speed unless acted upon by an outside force.
Almost. It will still have the same velocity in miles per hour, but as the radius of the circle increases the tangential velocity (radians/hour) will reduce. Since the radius of earth is approximately 4000 miles a difference of 10 will be small but still measurable.
 

diazona

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If there were no atmosphere, I think angular momentum conservation would take effect, and that would mean that the balloon's tangential velocity (mph) actually would decrease as it rose, in such a way as to keep L=mvr constant.

In reality, the atmosphere would act to increase the balloon's tangential velocity (in mph) such that its angular velocity (radians/hour) stays essentially the same. At least, any difference we would ascribe to prevailing winds.
 

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