Time to terminal

[atsaubrey 0]

I have a boss who is at odds with me on the time it takes to get to terminal from a stationary object. (helo/ballon) What amount of time would it take to get to terminal and about how many feet would it take. Only guesses here based on 120mph terminal.

"GOT LEAD?"

[NeedToJump 0]

The time required will differ depending on how you exit (belly to earth, HD, etc.) Are you assuming no wind resistance? (and therefore no terminal...)

Wind Tunnel and Skydiving Coach http://www.ariperelman.com

[JohnMitchell 14]

General answer is you that you reach 80% of terminal in about 8 seconds, then a few more seconds to hit 120 mph. A whole lot of variables in there. All that takes a little over 1000 feet, maybe 1200 feet or so.

[Harksaw 0]

120 miles per hour is equal to 176 feet per second. In zero air resistance a body accelerates 32 feet per second per second. Therefore, in zero air resistance a body reaches 120 mph in 5.5 seconds. In reality, it would take longer, how much longer I don't know.

__________________________________________________
I started skydiving for the money and the chicks. Oh, wait.

[quade 3]

Lift = ((((1/2rho)V^2)A)Cl)

Lift = 1/2 the density of the air in units called slugs times the velocity squared times the surface area times the coefficient of lift.

When lift = weight the object is at terminal velocity.

As I'm almost certain you're aware, terminal velocity is different for skydivers of different sizes and weights and is very much affected by their body position.

quade -
The World's Most Boring Skydiver

[NeedToJump 0]

Somewhat related question: does it take the same amount of time to reach terminal when exiting a hot air balloon vs. exiting an otter? It seems like it should since in both the vertical speed is zero and it is your vertical acceleration which causes you to reach terminal.

Wind Tunnel and Skydiving Coach http://www.ariperelman.com

[AggieDave 6]

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.

--"When I die, may I be surrounded by scattered chrome and burning gasoline."

[pilotdave 0]

Don't become a physics teacher...

Dave

[pilotdave 0]

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Lift = ((((1/2rho)V^2)A)Cl)

Lift = 1/2 the density of the air in units called slugs times the velocity squared times the surface area times the coefficient of lift.

When lift = weight the object is at terminal velocity.

Hehe whats the coefficient of lift of a human body belly to earth? I assume you're using the word lift because the force is pointed upward, but I'd call it drag (and coefficient of drag in the equation). Just words, no difference, but lift just sounds funny to me.

Dave

[darkwing 4]

OK, physics guy chiming in here. The question is vague enough to merit a vague answer. it takes 10 seconds, which is 1,000 feet. Sure that is a round, rule of thumb, but the devil is in the details. The "real" answer depends on variables including body position, body mass, body shape, exit altitude, and atmospheric conditions. Practically speaking there is no formula that you can plug the numbers in to and get a much better answer.

-- Jeff
My Skydiving History

[quade 3]

I guess I used lift because (remember the four forces?) lift opposes weight.

Anyway, that's my story and I'm stickin' to it.

As for the Cl of a human body, that would be determined the same way it is in almost all other cases -- put the body in the airstream, note the forces and then determine the Cl through experimentation. With our flexible bodies, you'd have to spend a lot of time in the wind tunnel.

quade -
The World's Most Boring Skydiver

[pilotdave 0]

Well a horizontal wind tunnel measures lift vertically and drag horizontally. You'd have to go to all the trouble of renaming the axes on your graphs!

Dave

[FrogNog 1]

Quote

Don't become a physics teacher...

Dave

While AggieDave's answer is disgustingly wrong from a sterile, vacuum-based physics point of view, it has some merits in a viscous medium.

If someone exits an otter in a perfectly neutral position and remains there - flying directly "down" on the relative wind - then they will fall vertically just as fast as someone who exited a stationary balloon at the same altitude at the same time. After X seconds, both jumpers will be at the same altitude, from exit all the way to opening. ("All other things being equal", of course; a standard scientific comparison assumption.)

But the person exiting the otter has airspeed they can use to fly at an angle from "down" on the relative wind. Just as RWers and birdmen can move "forward" while falling straight toward earth, they can move upward immediately after exiting a moving plane while falling horizontally. Even for the people who don't pop up entirely above the plane, they are countering gravity to some degree, and gravity is what accelerates them toward earth until their wind resistance matches the force of gravity on their mass. I believe someone else in this post gave the formula for velocity, which is the integration of acceleration, which is Force / Mass. The Force is the net of Gravity and the jumper's own flying force up or down, so they could achieve a higher or lower velocity in the same time as someone who lets gravity do all the work.

But for people who always fall straight down, it should be about the same amount of time. I was always told 11 seconds but 10 is close to 11.


Now, what about the definition of terminal velocity? Do we all agree that it's a stable velocity pointed straight toward earth? Because if we're just talking about terminal _speed_ (i.e. directionless), then we have a complication with people who exit planes above about 120mph. (Remember the 747 and military transport jumpers who say they _slow down_ to terminal. )

-=-=-=-=-
Pull.

[AggieDave 6]

Quote

While AggieDave's answer is disgustingly wrong from a sterile, vacuum-based physics point of view, it has some merits in a viscous medium.

Thanks.
If we jumped in a perfect conditions vacum like physic types use, one, we'd die, but two, I wouldn't need a really baggy jumpsuit either.

Step 1 of college physics: Assume the horse is a sphere and in a vacum...

--"When I die, may I be surrounded by scattered chrome and burning gasoline."

[Harksaw 0]

If we assume that terminal velocity is 120 mph and the acceleration due to gravity is 9.81 m/s^2, does anyone know if it is possible to determine the coeffecient of drag with this information?

__________________________________________________
I started skydiving for the money and the chicks. Oh, wait.

[NeedToJump 0]

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you have forward velocity with a plane, which arcs you at a given speed, speeding up as you begin downwards

Ummm, huh? I didn't think that horizontal speed had an effect on vertical speed...

Wind Tunnel and Skydiving Coach http://www.ariperelman.com

[quade 3]

I guess we could plug in ISA standard atmosphere (I'm not sure at what altitude we'd all want to use) and we'd have to come up with a "standard" skydiver (height, weight) falling in a "standard" position and then just guess that he was wearing a "standard" jumpsuit.

I guess the real answer is that you could do it for an anthropomorphic dummy posed and welded into a boxman (or some other) position, but humans seem to move around a bit.

quade -
The World's Most Boring Skydiver

[larsrulz 0]

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If we assume that terminal velocity is 120 mph and the acceleration due to gravity is 9.81 m/s^2, does anyone know if it is possible to determine the coeffecient (sic) of drag with this information?

Quade can stick to his little ideas, but line of flight of a body is directly down (yada yada....not for tracking...yada yada), since drag is the force resisting the "thrust" of gravity, then it isn't an issue of lift of any sort.

To answer your question, yes it is possible to determine the drag force of each particular human body/orientation (or your general belly-down human body of 120 mph). The jist is everything havinto resist the force of gravity can be clumped into this drag force. If you want to, you could even compute a Cd (like Cl that Quade mentioned, just this is the coefficient of drag). Now, you could do some interesting experiments to determine the change in terminal velocity as a body backslides (which can be seen as lift), so from there you could determine the skin friction/zero lift drag coefficient, which in turn could be used to determine the lift of a human body, but this would require some horizontally moving body position, such as a backslide or track.

I got a strong urge to fly, but I got no where to fly to. -PF

[Push 0]

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

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.

-- 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.

[mjosparky 3]

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120 miles per hour is equal to 176 feet per second. In zero air resistance a body accelerates 32 feet per second per second. Therefore, in zero air resistance a body reaches 120 mph in 5.5 seconds. In reality, it would take longer, how much longer I don't know.

You accelerate at the rate of 16 feet per sec., per sec.. Or 16 feet the first sec., 46 feet the second sec. and 76 feet per sec. the third sec.
In 12 sec. you will reach 174 feet per sec. which is just about 120 mph.

This is all approximate and your mileage may differ.
Sparky

My idea of a fair fight is clubbing baby seals

[kallend 1,623]

Quote

Quote

Don't become a physics teacher...

Dave

While AggieDave's answer is disgustingly wrong from a sterile, vacuum-based physics point of view, it has some merits in a viscous medium.

If someone exits an otter in a perfectly neutral position and remains there - flying directly "down" on the relative wind - then they will fall vertically just as fast as someone who exited a stationary balloon at the same altitude at the same time. After X seconds, both jumpers will be at the same altitude, from exit all the way to opening. ("All other things being equal", of course; a standard scientific comparison assumption.)

But the person exiting the otter has airspeed they can use to fly at an angle from "down" on the relative wind. Just as RWers and birdmen can move "forward" while falling straight toward earth, they can move upward immediately after exiting a moving plane while falling horizontally. Even for the people who don't pop up entirely above the plane, they are countering gravity to some degree, and gravity is what accelerates them toward earth until their wind resistance matches the force of gravity on their mass. I believe someone else in this post gave the formula for velocity, which is the integration of acceleration, which is Force / Mass. The Force is the net of Gravity and the jumper's own flying force up or down, so they could achieve a higher or lower velocity in the same time as someone who lets gravity do all the work.

But for people who always fall straight down, it should be about the same amount of time. I was always told 11 seconds but 10 is close to 11.

WRONG.

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

The reason is that drag goes as v^2, so is non-linear with velocity. This results in an interesting coupling between the horizontal and vertical components of the drag force, increasing the vertical drag component over what it would have been in the absence of horizontal velocity.

Jeez - even Newton knew that in 1676.

...

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

[Push 0]

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The reason is that drag goes as v^2, so is non-linear with velocity. This results in an interesting coupling between the horizontal and vertical components of the drag force, increasing the vertical drag component over what it would have been in the absence of horizontal velocity.

Jeez - even Newton knew that in 1676.

I think this is far from obvious to people not working with numbers all day long. It's not particularly obvious anyway. Newton was a lot smarter than all of us put together. There's no need to be mean about it.

-- 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.

[NeedToJump 0]

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The reason is that drag goes as v^2, so is non-linear with velocity. This results in an interesting coupling between the horizontal and vertical components of the drag force, increasing the vertical drag component over what it would have been in the absence of horizontal velocity.

Wow, very cool. Thanks for the explanation

Wind Tunnel and Skydiving Coach http://www.ariperelman.com

[cvfd1399 0]

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" 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 How a whuffo gonna tell me how to fall??

[kelpdiver 2]

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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.