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atsaubrey

Time to terminal

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The drag created by the horizontal motion creates some additional drag for the downard motion. At our nomal exit speeds it's small, but it's there.

The good Professor tends to speak in precise terms so while he is correct, the effect is essentially insignificant for us all things considered.
quade -
The World's Most Boring Skydiver

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I had the same argument with mine today. He kept telling me all objects fall at the same rate, and terminal velocity of all objects is 120 mph regardless of weight or surface are presented to the wind! I was like dude FREAKING listen to me I am an ANVIL I know how things fall relative to me and a 120 pounder will not fall the same rate as me when we are both belly to earth. He also stated that there is no reason for different type jump suits b/c "all you skydivers fall the same mph no matter how tight or baggy that suit is":S He then proceded to tell me that the different fall rates on my neptune are based on times, and that I was falling the same MPH, but had longer and shorter timed jumps that is what that is all about:S:S:S How a whuffo gonna tell me how to fall??



Tell him that if that were true, then using a parachute would be kind of pointless, wouldn't it?;)
Sky, Muff Bro, Rodriguez Bro, and
Bastion of Purity and Innocence!™

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It takes longer even in a neutral position to accelerate to a given vertical speed if you have some initial forward speed.



I don't see that as a given, Kallend. If you get your body in an inverted track while riding the wave, shouldn't that accelerate you downward in addition to gravity? Seems like body position could give you either positive or negative lift related to your exit point.



I didn't say it wouldn't. I said that IN A NEUTRAL POSITION that there is a coupling between horizontal and vertical components of drag. I said nothing at all about riding a wave or surfing the propblast.
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The only sure way to survive a canopy collision is not to have one.

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The drag created by the horizontal motion creates some additional drag for the downard motion. At our nomal exit speeds it's small, but it's there.

The good Professor tends to speak in precise terms so while he is correct, the effect is essentially insignificant for us all things considered.



Not that small, actually, especially immediately after exit when horizontal velocity is high.

You can easily demonstrate with ping pong balls. Use a slingshot to fire one horizontally while allowing the other to drop vertically, starting together at the same height. The dropped one reaches the ground first by a noticeable amount.
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The only sure way to survive a canopy collision is not to have one.

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I didn't say it wouldn't. I said that IN A NEUTRAL POSITION that there is a coupling between horizontal and vertical components of drag. I said nothing at all about riding a wave or surfing the propblast.



sorry, man. distracting day at work and I wasn't reading word for word, apparently.

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I didn't say it wouldn't. I said that IN A NEUTRAL POSITION that there is a coupling between horizontal and vertical components of drag. I said nothing at all about riding a wave or surfing the propblast.



sorry, man. distracting day at work and I wasn't reading word for word, apparently.



Work - the curse of the drinking class.
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The only sure way to survive a canopy collision is not to have one.

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I had the same argument with mine today. He kept telling me all objects fall at the same rate, and terminal velocity of all objects is 120 mph regardless of weight or surface are presented to the wind! I was like dude FREAKING listen to me I am an ANVIL I know how things fall relative to me and a 120 pounder will not fall the same rate as me when we are both belly to earth. He also stated that there is no reason for different type jump suits b/c "all you skydivers fall the same mph no matter how tight or baggy that suit is":S He then proceded to tell me that the different fall rates on my neptune are based on times, and that I was falling the same MPH, but had longer and shorter timed jumps that is what that is all about:S:S:S How a whuffo gonna tell me how to fall??



I can see what your problem is here - you are still trying to talk to whuffo's about skydiving. As soon as you stop talking to whuffo's about diving this problem should go away.


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I had the same argument with mine today..... How a whuffo gonna tell me how to fall??



This problem is easy to fix......stop talking to whuffo's about skydiving. Nod your head and say "yeah right" next time this happens and dont get drawn into a discussion - save it for the people in the sport.

ceers


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I tried some stuff with the equations and the results are working out pretty interesting. My computation went as follows. First, I assume that air density and area do not change on a low altitude BASE jump, which is fairly reasonable. Then, I define my Cd' = A*r*Cd / 2 to be my new coefficient of drag, so that I don't have to worry about things that we depend linearly on that I assumed to be constant, to give a very simple equation for drag:

D = Cd' * v^2

Note that I will use V for terminal velocity and v for arbitrary velocity. At terminal we have

W = D => mg = Cd' * V^2 => Cd' = mg / V^2

Now, from Newton's Second Law:

ma = mg - Cd' * v^2 = mg - (mg / V^2) * v^2, so

dv/dt = g - (g v^2) / V^2 = g (1- v^2/V^2)

Integrating both sides with respect to t gives:

v = g (t - v^3/ (3*V^2))

Finally, solving for t gives:

t = v/g + v^3 / (3*V^2)

This equation gives me 23 seconds to reach 53.33 m/s with terminal velocity 53.33 m/s. This is not particularly surprising, since actually reaching the complete terminal velocity should take a long time. For v = 40 m/s with V = 53.33 m/s I get 11.6 seconds. Consider the protrack graph in this thread: http://www.dropzone.com/cgi-bin/forum/gforum.cgi?post=390101;search_string=protrack%20graph;#390101.
The equation I derived predicts this graph with more than reasonable accuracy. Yes, I know that the poster is asking about inaccuracies in the graph:P

Edited to add: 53.33 m/s = 120 mph. If you want to plug numbers into the above equation, you HAVE to do it in m/s because of the units of g. Here is a units converter.


Your solving of the differential equation is quite wrong which should be obvious if you consider the behaviour after reaching terminal velocity.

The correct solution to the equation is a hyperbolic tangent function. Fitting the constants to the assumtion that initial acceleration is 32 feet per second squared and terminal velocity is 176 feet per second we get v=176*tanh(t/5.5). By this formula you should never reach terminal velocity (and cosequently never exceed it), 90% should be reached after about 8 seconds, 95% after about 10 seconds, 97.5% after about 12 seconds.

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This problem is easy to fix......stop talking to whuffo's about skydiving.



Great, 43 jumps and you're already too good to talk to whuffos... Just because you skydive doesn't make you a better person than someone who doesn't. Please don't fall into the trap of thinking that you are... :S
Wind Tunnel and Skydiving Coach http://www.ariperelman.com

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Yep, I know. I used the "integrate both sides of an equal sign to get something approximate" method of solving DEs:D Plugging in e^ax would be the right way to go about it, yeah. Integrating both sides of an ODE sometimes gives a polynomial solution that is accurate up to some terms of the Taylor expansion of the actual solution. I was hoping this could give a similar behaviour when v/V < 1. Of course the asymptotic behaviour would be wrong.

-- Toggle Whippin' Yahoo
Skydiving is easy. All you have to do is relax while plummetting at 120 mph from 10,000' with nothing but some nylon and webbing to save you.

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Not quite, you have forward velocity with a plane, which arcs you at a given speed, speeding up as you begin downwards. With a balloon there is no arc, obviously so you have to "earn" every bit of speed you get.

I'n gonna say you're wrong. I know you are a shooter, too, and know that the vertical speed of the bullet (drop) is independent of the horizontal speed. A skydiver jumping from a balloon would accellerate and fall at the same rate as a jumper leaving a passing airplane. The only difference would be the jumper from the plane would also move across the sky from the forward throw of the plane.

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This problem is easy to fix......stop talking to whuffo's about skydiving.



Great, 43 jumps and you're already too good to talk to whuffos... Just because you skydive doesn't make you a better person than someone who doesn't. Please don't fall into the trap of thinking that you are... :S



I don't think that not talking to whuffos about skydiving implies you think you're better than them. I am personally stubborn and love to blab about jumping to everyone even if I get made fun of or called insane. BUT, I do understand how tiring the same old remarks get, and I can see why someone might make a personal decision to stop talking about skydiving around whuffos. If I made that decision it wouldn't be because I think I'm better than them, but because I'm happier without those discussions.
www.WingsuitPhotos.com

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Fine, I'll do it. Nonlinear nonhomogeneous first order DE, yay. It's of the form df/dx ~ 1-f^2, so as you suggest, set v = a tanh (ct). Then dv/dt = a (1 - tanh^2 (ct)) c. Since tanh^2 (ct) = v^2 / a^2, we have dv/dt = a (1 - v^2 / a^2) c = (a - v^2/a) c. Setting a = V and c = g/V we are done. Solution is:

v = V tanh (gt/V)

Plugging in some numbers, the lazy method gave an error of about a factor of 2.

Edit: Which is very interesting considering that protrack graph. Hrm. Must have been that SAS measurement, instead of TAS.

-- Toggle Whippin' Yahoo
Skydiving is easy. All you have to do is relax while plummetting at 120 mph from 10,000' with nothing but some nylon and webbing to save you.

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Yep, I know. I used the "integrate both sides of an equal sign to get something approximate" method of solving DEs:D Plugging in e^ax would be the right way to go about it, yeah. Integrating both sides of an ODE sometimes gives a polynomial solution that is accurate up to some terms of the Taylor expansion of the actual solution. I was hoping this could give a similar behaviour when v/V < 1. Of course the asymptotic behaviour would be wrong.



It doesn't seem as if you know what you're writing about.
...

The only sure way to survive a canopy collision is not to have one.

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How much of an effect is this drag coupling?

P.S. Sir Issac Newton knew a lot of stuff I still don't know.:$



D = - k V^2 = -k (Vx^2+ Vy^ 2 )
(V is velocity vector, Vx, Vy are its components, similarly for the drag force D)

resolving in x, and substituting for cosine of the angle:

Dx = - k (Vx^2+ Vy^ 2) Vx / sqrt(Vx^2+ Vy ^2 )

= - k Vx sqrt(Vx^2+ Vy ^2 )

Similarly for Dy = -k Vy sqrt(Vx^2 + Vy^2)

so let y = vertical and x = horizontal, if Vx >0 then the vertical Dy is greater than it would be if Vx = 0

It will add about 1 second to the time to fall the first 1000ft if the jump plane is going 90kt, as compared to a balloon. Once you're off the hill the effect disappears.
...

The only sure way to survive a canopy collision is not to have one.

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Whatever you say, professor. The idea is that in some very special cases, you could pretend that the Taylor expansion of the solution has the first couple of terms to be what you want to make the DE essentially trivial. It was a homework exercise, to find cases where the error is bounded. I haven't dealt with DEs ever since, so it's possible that this approximation is something the professor came up with as a homework exercise. Anyway, I already found the solution.

-- Toggle Whippin' Yahoo
Skydiving is easy. All you have to do is relax while plummetting at 120 mph from 10,000' with nothing but some nylon and webbing to save you.

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