formula for height versus time? (for freefall)

[mixedup 0]

Anyone know a link to a freefall formula? What I'd be interested in is one that is like:

height = function of (time)

But also has a way to change a parameter/constant or whatever to change the formula for basic belly flying, head down, free fly etc.

So the formula would have to take account that in the first 1000 feet it takes longer too, as speed builds up.

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[LouDiamond 1]

Like THIS? Click on "Formulae" at the top left of the screen if you want to see the math.

"It's just skydiving..additional drama is not required"
Some people dream about flying, I live my dream
SKYMONKEY PUBLISHING

[mixedup 0]

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Like THIS? Click on "Formulae" at the top left of the screen if you want to see the math.

thanks but unfortunately these don't seem to take account of wind resistance/terminal velocity or style of skydive

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[mixedup 0]

actually found what seem to be reasonable formula's at http://en.wikipedia.org/wiki/Free_fall in the "Uniform gravitational field with air resistance" section.

Just need to find out CD is the drag coefficient for different styles of skydive now I think

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[LouDiamond 1]

what are you trying to do?

"It's just skydiving..additional drama is not required"
Some people dream about flying, I live my dream
SKYMONKEY PUBLISHING

[councilman24 37]

Since 'terminal' velocity depends on body position, size, mass, jumpsuits, etc; and body positions and jumpsuits vary a WHOLE lot more than they did in 1980 no one 'table' will till you much.

speed/time = distance

You need to decide if your going 40 mph or 250 mph vertical.

Then add in time to terminal from exit.

I'm old for my age.
Terry Urban
D-8631
FAA DPRE

[mixedup 0]

agreed - what do you think the terminal speeds are for the various forms of skydiving - so far I've found....

* belly flying - 55m/s = 122miles/hour
* freeflyer - 90m/s = 200miles/hour
* headdown - ??

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[ibx 2]

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* headdown - ??

Headdown is one of the freelfy positions.
If you mean speed skydiving, the world record is 527km/h I think. This is not the max speed, but an average over 1000m.

You can add about 10 sec. until you reach terminal...

Freeflying can go from about 250km/h to about 280km/h. It's gotten slower since everybody has tunnel time

[mixedup 0]

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Freeflying can go from about 250km/h to about 280km/h. It's gotten slower since everybody has tunnel time

so people want to fly slower here? I heard from my instructor they try to fly faster for better control but that was probably for belly flying I guess...

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[ibx 2]

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so people want to fly slower here? I heard from my instructor they try to fly faster for better control but that was probably for belly flying I guess...

Yeah he probably meant belly flying. I think with freeflying it has more to do with the techniques you learn in the tunnel... The first tunnels where slower than many are today, this made it very hard for people to take in freefly positions so they developed ways to fly slowly.

[PWScottIV 0]

There's not an individual formula that I'm aware of that will solve for speed or height as a function of time for a whole skydive. You would have to break it down into at least two calcs. Even then, your results won't be exactly accurate for the beginning of the skydive due to the forward speed of the aircraft, unless you're talking about a balloon jump, and then that would be easier to calc.

In the beginning of the skydive the jumper has a linear momentum in the direction of the aircraft's flight. There is drag force in the direction of the relative wind (which begins roughly parallel to the ground and transitions to roughly perpendicular, and aligned with gravity). And finally there is the force of gravity. This transition period is known as "the hill". During this period, depending on the orientation of the skydiver, there can be some amount of lift generated from the horizontal component of the skydiver's velocity, which can complicate the calc... I'm not really sure how much that is, but it might be negligible for your purposes.

If you were to look at a balloon jump, accounting for the velocity with respect to time is easier. And like I mentioned, it might not be a whole lot different than a skydive out of a moving aircraft. In this type of jump, there is only one initial force, gravity in the vertical direction, and the velocity is zero, so the drag force is zero. Assuming the skydiver is stable in only one position, then you can use standard cals for a falling object (http://www.grc.nasa.gov/WWW/k-12/airplane/falling.html). But you don't know the Cd... What you do know, or can make assumptions about, is the terminal velocity in a particular body position, for a typical skydiver. Calculating a "real" Cd for a skydiver at more than just an instant would be nearly impossible... Way too many variables. However, you can back calculate from your assumed terminal velocity... Which again, can vary widely based on more variables than anyone would want to think about.

Assuming again that the skydiver is in a stable position, with a constant Cd, once the skydiver reaches terminal velocity (which is only theoretical), he/she actually starts slowing down. This is because, as the skydiver falls, the density of the atmosphere increases, which increases the drag force until the skydiver ends the freefall portion of the skydive. Furthermore, if you care, the density of the air is also affected by temperature, which typically increases as altitude decreases... That slightly counteracts the first density change I mentioned which is due to the weight of the atmosphere compressing itself.

Eventually, for a belly-to-earth skydive, you should find that it takes about 10 seconds to reach terminal velocity and at that point the skydiver has lost about 1000ft of altitude. Once terminal (theoretical) is reached, then the skydiver's altitude decreases by about 1000ft every 5.5 seconds for the remainder of the skydive.

Let me know if you have any other questions.

Gravity Waits for No One.

[mixedup 0]

good explanation - thanks - for the purpose of some mucking around I found the formula's at http://en.wikipedia.org/wiki/Free_fall in the "Uniform gravitational field with air resistance" section wasn't too bad. That is, it covered at least the difference between the initial 10 seconds until terminal speed is reached ok.

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[JackC1 0]

To do it properly you need to solve two coupled differential equations but if you ignore the fact that the plane is flying horizontally (which slows the acceleration downwards due to the horizontal drag), you can simplify the problem to a single differential equation. The answer you then get looks like this.

[PWScottIV 0]

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To do it properly you need to solve two coupled differential equations but if you ignore the fact that the plane is flying horizontally (which slows the acceleration downwards due to the horizontal drag), you can simplify the problem to a single differential equation. The answer you then get looks like this.

If you separate vertical forces from horizontal, and only look at the vertical, then you're left with gravity, the vertical drag force component, and vertical lift force (if any). It's not horizontal drag force that affects the vertical acceleration component... It's the vertical lift force. If the person was tucked into a symmetric ball position, then the effect of the horizontal velocity would be negligible. Also, the equations you presented don't take into account the variation of air density as a function of altitude... An effect that can make a skydiver's velocity vary as much as 10mph.

Gravity Waits for No One.

[JackC1 0]

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It's not horizontal drag force that affects the vertical acceleration component... It's the vertical lift force. If the person was tucked into a symmetric ball position, then the effect of the horizontal velocity would be negligible.

Sorry, you're wrong. The differential equations you need to solve are:

dvy/dt = g - 0.5*C*p*(vx^2 + vy^2)/m
and
dvx/dt = - 0.5*C*p* (vx^2 + vy^2)/m

You cannot split these equations up into x and y components so the drag caused by the horizontal velocity (vx) actually slows the acceleration downwards (dvy/dt). It makes a difference, even for a bowling ball although lift from body position would probably make a bigger difference.

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Also, the equations you presented don't take into account the variation of air density as a function of altitude... An effect that can make a skydiver's velocity vary as much as 10mph.

Then write the density (p in the above) as a function of altitude and solve the equations again. You'll probably need to do that numerically. I doubt an analytical solution would be possible.

[PWScottIV 0]

Where did you get your equations? I'm not necessarily saying they're wrong, just confusing to me... Breaking down the first eqn doesn't make sense how that could be right. if it is, it seems like it would have more to do with the linear momentum of the skydiver and not the drag. I could see how this eqn would make sense if it described the absolute velocity, but not if it's supposed to be for just one component (x or y).

dVy/dt = ay = = Fy/m = g - (0.5*C*p*Vx^2)/m + (0.5*C*p*Vy^2)/m

Gravity Waits for No One.

[godfrog 2]

perhaps in consideration should also be where on the earth you are applying these formula as gravity is not constant and consistant across the earth.

Experience is a difficult teacher, she gives you the test first and the lesson afterward

[PWScottIV 0]

lol, I assume that was tongue in cheek? If not, I guess it depends on what the end use of the calculation is... If it was for a math or engineering class, I would at least want to know what the variables were to develop a valid and complete list of assumptions. I would hope that any instructor worth their salt would at least appreciate their students being thorough about examining a problem.

Gravity Waits for No One.

[buff 0]

It is a Saturday night and the only thing y'all have proven is that Shah and you guys will never get laid and this may be the end times of the boogie season and mankind as we know it.

It's called the Hillbilly Hop N Pop dude.
If you're gonna be stupid, you better be tough.
That's fucked up. Watermelons do not grow on trees! ~Skymama

[PWScottIV 0]

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It is a Saturday night and the only thing y'all have proven is that Shah and you guys will never get laid and this may be the end times of the boogie season and mankind as we know it.

1. I get laid whenever I choose... Basically daily.
2. I live in California... We skydive and have boogies year-round.

Gravity Waits for No One.

[JackC1 0]

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Where did you get your equations? I'm not necessarily saying they're wrong, just confusing to me... Breaking down the first eqn doesn't make sense how that could be right.

Sorry, they are wrong. I forgot to resolve the drag forces into their x & y components. These are better, I think.

dvy/dt = g - 0.5*C*A*p*vx*sqrt(vx^2 + vy^2)/m
and
dvx/dt = - 0.5*C*A*p*vy*sqrt(vx^2 + vy^2)/m

[PWScottIV 0]

Have a source for that? Couldn't find it in my fluids books.

Gravity Waits for No One.

[JackC1 0]

I'm sure it's written down somewhere, although I forget where. I just wrote down Newtons second law for the system. The bit that causes the mixing (and the confusion) is the drag term which is -0.5*C*A*p*V^2. Drag doesn't care what direction the wind comes from, it just goes with the magnitude of the velocity squared, where V^2 = vx^2+vy^2. You then resolve this into x & y components by multiplying by the cos or sin of the angle, which is vx/sqrt(vx^2+vy^2) and vy/sqrt(vx^2+vy^2).

If you state in your boundary conditions that the initial horizontal velocity vx=0, then the whole lot simplifies to:

dvy/dt=g-0.5*C*A*p*vy^2/m

Which is just a=F/m where F is gravity (g) and drag (-k*v^2)

[kallend 2,146]

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Where did you get your equations? I'm not necessarily saying they're wrong, just confusing to me... Breaking down the first eqn doesn't make sense how that could be right. if it is, it seems like it would have more to do with the linear momentum of the skydiver and not the drag. I could see how this eqn would make sense if it described the absolute velocity, but not if it's supposed to be for just one component (x or y).

dVy/dt = ay = = Fy/m = g - (0.5*C*p*Vx^2)/m + (0.5*C*p*Vy^2)/m

Because drag goes with v^2 (at the Reynolds Numbers involved) the x and y equations ARE coupled.

You don't ALWAYS have to "get" equations from somewhere. You can work them out for yourself.

...

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

[PWScottIV 0]

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Where did you get your equations? I'm not necessarily saying they're wrong, just confusing to me... Breaking down the first eqn doesn't make sense how that could be right. if it is, it seems like it would have more to do with the linear momentum of the skydiver and not the drag. I could see how this eqn would make sense if it described the absolute velocity, but not if it's supposed to be for just one component (x or y).

dVy/dt = ay = = Fy/m = g - (0.5*C*p*Vx^2)/m + (0.5*C*p*Vy^2)/m

Because drag goes with v^2 (at the Reynolds Numbers involved) the x and y equations ARE coupled.

You don't ALWAYS have to "get" equations from somewhere. You can work them out for yourself.

Remembering back to fluids, the equations we worked with for lift and drag of an airfoil were what I was thinking of... Those equations had been simplified to only consider drag force opposite to the flight direction. I guess it just seems counter-intuitive that drag, resolved as a force, would effect an object from a direction perpendicular to the flight path.

Gravity Waits for No One.