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jigneshsoni

Simple Physics Question

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I have been thinking about this since a few days. I would like to know what is the relation between the distance covered in the air at a certain height AND distance covered on ground.

Eg: I would like to know if I am at point A in the air which is 5000 feet altitude. If I move towards point B in the air which is 100 feet North from Point A. What I am actually doing with respect to a viewer on the ground?

I have a feeling that when you move just a small distance in the air, you are actually moving quite a big distance with respect to ground. Is it true or is it just how it feels OR is it just how I think which is totally wrong from reality?

If anybody could send me a link which would explain me this theory in more details, that would be great.

Thanks
Jigs

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i am not a physics person or anything, so i will let someone else answer this.

but i will encourage you to read _the parachute and its pilot_, which was SO helpful for me and talks about lots of these principles. brian germain. i stole a copy from one of my coaches :) still need to give it back now that i am finished...
life is either a daring adventure or nothing at all.
(helen keller)

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if i understand your question, you have it backwards.

at higher altitudes, you have to travel farther to reach any set radial from the center of the earth.

or, the closer you are to the center of the earth, the less you have to travel to do the same thing.

the circumference of a circle with a smaller radius, in relation to a circle with a larger radius, is smaller than the larger circle.

does that make sense?
"Don't talk to me like that assface...I don't work for you yet." - Fletch
NBFT, Deseoso Rodriguez RB#1329

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I does make sense a little bit. But here is an example, Please comment.

When you look straight down while freefall from 10,000 feet, you see an area of around 10,000 sq feet as small as a 1 sq feet. Lets say you see that 1 Sq feet area at the center when you look down. Now when you move forward, you moved like say 10 feet.
So now you see at the center another area which is 1 sq feet, but ofcouse it is 10 feet away. If we keep falling, will we not fall at the new point??

Won't the new point be actually some point in the new area which is actually 100,000 feet away from the original area?

I hope what I am saying makes sense, Please comment.

Thanks
Jigs

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This really doesn't have much to do with physics (optics, maybe).

The explanation is pretty simple - things look small when they are further away.

If your example was valid, it would also work across horizontal distances, in addition to freefall (vertical distances). An analog would be saying that you can build a perfectly straight highway that is 3 miles long. If you stand at one end of the highway and your friend stands at the other, so that you are both 3 miles apart, on the same side of the road, you will appear very small to each other. By your logic, if you walk 20 feet to cross the road, you have actually walked much further because he appears so small to you. In other words, if you walked the 20 feet to cross the road, then walked the 3 miles to the other end of the road where he was, you should somehow be more than 20 feet away from him now. It doesn't make sense.
www.WingsuitPhotos.com

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i see what you mean and the answer to this question:

Quote

Won't the new point be actually some point in the new area which is actually 100,000 feet away from the original area?



is no.

no matter where you move to in the atmosphere, if you are falling, you are falling toward the center of the earth. therefore, my original circle example still holds true. if you move 10 ft laterally in the air at 10,000 feet, you have moved a smaller distance across the ground.

you can calculate these distances by using the diameter of the earth and another diameter that is twice the given altitude of the skydiver plus the diameter of the earth.
"Don't talk to me like that assface...I don't work for you yet." - Fletch
NBFT, Deseoso Rodriguez RB#1329

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For all practical purposes, when you jump from an airplane, the Earth is flat. So when you move 100ft horizontally in the air, your projection on the ground also moves 100ft.

The illusion you have "that when you move just a small distance in the air, you are actually moving quite a big distance with respect to ground", is similar to the illusion that moon is bigger when it's close to horizon. Our eye perceives the sky as a "ceiling" parallel to the ground, and thus stellar objects near horizon appear to be farther from us. Since their angular size does not actually change, they appear to be physically bigger near horizon. But if you take a photograph, you'll see the size is the same.

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When you look straight down while freefall from 10,000 feet, you see an area of around 10,000 sq feet as small as a 1 sq feet. Lets say you see that 1 Sq feet area at the center when you look down. Now when you move forward, you moved like say 10 feet.
So now you see at the center another area which is 1 sq feet, but ofcouse it is 10 feet away. If we keep falling, will we not fall at the new point??

Won't the new point be actually some point in the new area which is actually 100,000 feet away from the original area?



No. Although the large area of ground you are looking at will appear to be much smaller than it is, it will also appear to move much slower than it really does. If you move ten feet forwards at 10,000ft up then I guarantee that if you look at the ground you will not appear to have moved at all. Honestly, your sight picture will not appear 1 millimeter different.

Like The111 said, what you are suggesting just does not make sense.
Do you want to have an ideagasm?

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I don't know if this is what you're looking for, but this kind of explains something. Draw a right triangle and use the pythagorean theorem. They'll be 3 points, a spot on the ground, your original spot 5000 ft above and your new spot 100 feet over. So 5000 squared + 100 squared = 25,010,000. So, the square root of 25,010,000 is 5,001. Even though you moved 100 feet up there you are only 1 foot further from the original point on the ground in distance.

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They'll be 3 points, a spot on the ground, your original spot 5000 ft above and your new spot 100 feet over. So 5000 squared + 100 squared = 25,010,000. So, the square root of 25,010,000 is 5,001. Even though you moved 100 feet up there you are only 1 foot farther from the original point on the ground in distance.



I like this answer almost as much as the one about "you're all on crack" but the original question, I think, had to do with what the perception of a viewer on the ground would be (???). Because the actual distance to the falling parachute would differ by such a small amount per the calculations above, there sure could not be much optical illusion effect caused by a change in distance between observer and parachute as the parachute glides along. The one variable that was not mentioned here was that the canopy does not remain at same altitude--it is descending (assuming no brakes) at about 3:1 glide ratio. If it moved 100 feet laterally it would sink simultaneously by only about 30 feet, a minuscule part of 5000 feet, so there is no optical illusion due to the chute "getting closer" and seeming larger thereby to the ground observer. Assuming no winds at 5000 feet, at typical speeds of student canopies, about 12 mph ???, the 100 foot "lateral" movement would take about 6 seconds. I believe the ground observer standing directly under the time zero position would see a canopy at time zero and the same canopy at time zero plus 6 seconds and they would look exactly the same size. To the naked eye, there would be some appreciation of slight movement over the 6 seconds, but not much with the canopy one mile up in the sky. As the canopy got closer to earth the same 100 foot forward travel would become progressively more obvious to the ground observer. Really close to the ground (e.g. at 300 feet) a 100 foot lateral movement would look like a 100 foot lateral movement at 300 feet altitude--huge.

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As the canopy got closer to earth the same 100 foot forward travel would become progressively more obvious to the ground observer. Really close to the ground (e.g. at 300 feet) a 100 foot lateral movement would look like a 100 foot lateral movement at 300 feet altitude--huge.



Isn't this what my calculation proves??? You can't notice 1 ft. when it's 5,000 feet away. But if you want to use the 300 foot scenario, that's a difference of 16 feet in my calculation. You may notice 16 feet when it's only 300 feet away.

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Isn't this what my calculation proves??? You can't notice 1 ft. when it's 5,000 feet away. But if you want to use the 300 foot scenario, that's a difference of 16 feet in my calculation. You may notice 16 feet when it's only 300 feet away.



This is obvious. It's why swoopers swoop.

The original poster however, seemed to be saying that perceived distance traveled somehow equated to actual distance traveled.
www.WingsuitPhotos.com

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I have a feeling that when you move just a small distance in the air, you are actually moving quite a big distance with respect to ground. Is it true or is it just how it feels OR is it just how I think which is totally wrong from reality?



I'm not 100% sure what you're asking... but try this. Watch a big way or a formation from the ground. It's quite easy to do if you concentrate on them all the way from exit... so long as your eyesight is ok.

They'll break off and track away from each other at arround 4 or 5 grand or so. Watch how far they go from one another. You can really shift when you track hard and watching from the ground can help you figure out just how far you have the potential to move up there.

Athough as I said, I'm not 100% that's exactly what you're asking... I hope it will help.

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For all practical purposes, when you jump from an airplane, the Earth is flat. So when you move 100ft horizontally in the air, your projection on the ground also moves 100ft.



This is my take also. While it is technically true that you are always falling towards the centre of the earth, from 10K the difference is negligible. Consider the flock that recently flew from Gibraltar to Africa. The horizontal distance that they flew corresponds pretty much with the distance across the face of the earth that they travelled. (Clearly they flew a bit further in actual distance covered given that they were descending at an angle, but for the purposes of this discussion, that is of little consequence.)



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Diagram is great, the distance moved would be less on the ground, only problem is the dif between the distance from the center of the circle (core of the earth) and the surface of the earth VS the distance between the core of the earth and 10k feet agl is such a tiny fractional diference you can ALMOST neglect it

Good Judgment comes from experience...a lot of experience comes from bad
judgment.

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The big problem is if jump run is directly reverse of the rotation of the earth
You will be falling with a slightly slower rotation than the earth for a while:D


"be honest with yourself. Why do I want to go smaller? It is not going to make my penis longer." ~Brian Germain, on downsizing

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I’ll have a go at this one.

Using your scenario lets take two people. One is at 100ft looking straight down, the other is at 10000 ft also looking straight down. Both of these people have a field of view that takes in 90° (because it makes the maths simple) in one direction on the ground.

The unlucky chap at 100 ft looks down at a fence he knows is 100 ft long. The fence takes up 50 % of his field of view in that direction. (The total length he can see in that direction is 2 x 100 x tan(45) = 200 ft).

The other fellow (who paid more) is enjoying a view from 10000 ft of the same fence. His field of view encompasses 20000 ft (2 x 10000 x tan(45)) in the direction of the fence. Thus the 100 ft fence only takes up 0.5 % of his total field of view. These two guys are shown in the attached picture as (1) and (2).

Now the plucky skydiver at 10000 ft jumps. Skilful as he is he moves 10 ft horizontally without losing any height (Lets just assume very little height is lost at least). The fence is still taking up 0.5 % of his field of view, but now 60 ft (0.3 %) of it is one side of the centre of his view and 40 ft (0.2 %) of it is the other side.

According to your assumption; as he now descends his field of view continues to be 20000 ft. So when he gets to 100 ft and is kissing his a$$ goodbye he looks down and sees our fence as 50 % of his view. Crikey, that fence must be 10000 ft long he thinks. Well I moved 10 % of its length earlier; I must have travelled 1000 ft up there. Fuzzy logic I think you’ll agree.

In reality the clever skydiver will realise he can now only see 200ft in total, so the fence is 50 % of 200ft = 100 ft and he moved 10 % of it, 10 ft. Not that it’s going to do him much good at 100 ft in freefall.
_________________________________________________

The effect of the curvature of the surface of the earth, or indeed heading towards the centre of it, is indeed negligible as others have said. A quick google reveals the radius of the earth to be about 2090000 ft. Although obviously we know that not only is the earth not a perfect sphere it is also quite hilly, this seems like a good number to use.

A trivial calculation shows 10ft movement (parallel to the earths surface) at 2100000 ft from earths centre is 7.5788x10^-5 % of the total circumference of a sphere of this radius. Now 2090000 ft from the centre of the earth (on the ground) 7.5788x10^-5 % of the circumference equates to 9.95 ft. A 0.05 ft difference.
__________________________________________________

I hope all my talk of fences has helped not hindered ;)

James

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So if we take hypothetical or real numbers and plug them in we can say:


a C
(for the picture of this go to my last post)

a=centerline straight up
b=centerline to landing point
c=total distance between landing and exit from the air
d=total distance between landing and exit from the ground

c/a=d/b

Plug in the known

Lets just say c is 200 feet traveled on the horizontal while in FF

The earth is 36,010,000 feet giving is A
B is 26,000,000

200/360,010,000 = d/x
5.555e = d/36.000,000
d = 199.94

So if we travel a distance of 200 feet in the air we travel a distance on 199.94 feet on the ground.

This is FUN! :)

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