Physics of Skydiving

[SeanCulver 0]

I am new to the sport and I am currently reading the book "skydivers handbook" In this book the author states that Weight and Drag are both Ratios that determine speed. I am confused I remember back in middleschool doing experiments while dropping two objects one extremely heavy and one that is light and both hitting the ground at the same time. Newtons law of gravity maybe... anyways it seems like that would apply towards skydiving as well which means weight is not a factor but simply the amount of surface area that the diver is. Anybody wanna clue me in? Thx

Sean

[Cajones 0]

There are a few other factors your experiment in middle school couldn't/didn't take into account.

The main factor to look at is air friction. We skydive in the atmosphere (not in a vacuum, normally), so the air itself is a big influence in determining the maximum speed an object can obtain. This maximum speed, generally referred to as "terminal velocity" is the balance of the acceleration of gravity pulling the object toward the center of the Earth, and the opposing friction of the object moving through the medium (air). The simplest illustration is a feather vs. a bowling ball. The feather has a great deal of influence from air friction due to its tremendous surface area vs. its mass. The bowling ball on the other hand, represents the way some of us cut through the air as we skydive - lots of mass, with little to slow us down, except - hopefully - a fully functioning parachute.

The laws of physics are strictly enforced.

[quade 3]

On the Moon, where there is little atmosphere, a feather and a hammer would fall at the same rate. As a matter of fact, on the EVA for Apollo 15, they demonstrated just that for kids watcing on TV. HERE is QuickTime movie of the demonstration.

On the Earth, the two would fall at very different rates because of the amount of drag versus the amount of potential energy (weight) each item has to overcome that drag.

While the hammer has a larger surface area creating more drag than the feather, it also has much more energy to overcome that drag.

This is also why if you have two parachutes of the same size with different weights under them, they'll reach the ground at different times.

quade -
The World's Most Boring Skydiver

[winsor 187]

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I am new to the sport and I am currently reading the book "skydivers handbook" In this book the author states that Weight and Drag are both Ratios that determine speed. I am confused I remember back in middleschool doing experiments while dropping two objects one extremely heavy and one that is light and both hitting the ground at the same time. Newtons law of gravity maybe... anyways it seems like that would apply towards skydiving as well which means weight is not a factor but simply the amount of surface area that the diver is. Anybody wanna clue me in? Thx

Sean

FWIW, to really get into the issues you raised would take the better part of a semester.

Weight is the force on a object due to gravitational attraction. The short form is:

W = mg

though both weight and g are vector quantities and m, mass, is a scalar multiplier. The magnitude of g is a function of altitude, but we treat it as constant for the sake of simplicity.

L/D is your lift to drag ratio. It is treated as a constant for a particular parachute, though it isn't really.

Surface area is one factor in drag, and it isn't all that simple. We lump relevant factors together in ballistics and come up with a coefficient which relates speed to drag in a given flow regime (here we're talking way subsonic).

The two objects dropped hit the ground at the same time only if they have the same ballistic coefficient and/or are dropped in a vacuum. All that says is that the magnitude of g (see above) is the same for both.

In any event, I wouldn't ponder too greatly on these issues from a theoretical standpoint, unless you feel like undergoing a rigorous treatment of the subject. If you do, prepare to unlearn an awful lot.


Blue skies,

Winsor

[bmcd308 0]

Your knowledge of physics so far ignores air resistance - a force that acts in a direction opposite the velocity vector of the falling object. Remember in school how your textbooks would always say "in a vacuum" or "ignoring the effects of air resistance"? Well, at skydiving speeds, you can no longer ignore it, so you wind up with a more complex motion formual than x=(1/2)*at^2. Now, you can't substitute g for a. In reality, a itself changes with velocity.

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www.jumpelvis.com

[billvon 2,454]

At terminal velocity you're not accelerating or decelerating, which means all forces are in balance. Your weight (W) pulls you towards the earth; you can use the universal gravitation formula, but for our purposes, you can just claim that a weight of W causes a downward force of W. Drag has to exactly oppose that, and the formula for drag looks something like A*Cd*V^3, where A is surface area, Cd is coefficient of drag, and V is velocity. So at terminal W = A*Cd*V^3.

Now you can plug numbers in. Want to fall faster? You have three choices. Increase your weight (add a weight belt for example) decrease surface area (pull your arms and legs in) or change your coefficient of drag (use a nylon suit instead of a cotton one.) If you do any of those things, you will accelerate until the above equation once again balances. You can also increase the mass of the earth or increase the gravitational constant, but those are hard to do.

[Frodo 0]

So at terminal W = A*Cd*V^3.At low speeds, air resistance is almost proportional to speed; as you go to higher speeds, the propportionality becomes square (Resistance ~ V^2); this is what I'd read in a good book on physics ; maybe at even higher speeds the resistance becomes proportional to V^3, as you're saying. I'm not sure. I just know that there is no single formula for air resistance, so you have to use approximations at some point.

[billvon 2,454]

>At low speeds, air resistance is almost proportional to speed . . .

At low speeds, power required to move a _car_ is proportional to speed, but that's because rolling resistance dominates. I don't think air resistance scales that way.

[quade 3]

Drag increases at a rate equal to the square of the factor of the increase in speed (for sub-sonic airspeeds and all other things being equal).

If you had 5 pounds of drag at 10 mph, then at 20 mph you'd have 5*(2^2) pounds of drag or 20 pounds of drag.

quade -
The World's Most Boring Skydiver

[SeanCulver 0]

The reason i was curious is because it discussed usuing wieght to increase the speed in which you dive at. I remember those experiments and wondered if that would really effect the rate at which you decended. Anyways you all have been really helpful. I think now have a better understanding of the subject.

Thx

Sean

[NoShitThereIWas 0]

Just to add to what Billvon said, there is also another formula for the amount of drag (resistant) force exerted on a skydiver’s body.

R= ½ (D)(p)(A)(v^2). Where R is the drag (Resistant) force, D is the drag coefficient, p "rho" is the density of the medium (air), A is the cross-sectional area of the falling object and v is the velocity of the object.

If velocity or cross-sectional area is increased, the drag force will also increase. When the drag force acting on a skydiver equals the force of gravity, the skydiver has a net force of 0 Newtons and has reached terminal (max) velocity for his/her body position.

Also as you go higher in altitude the air gets "thinner" and its density decreases. Therefore, the resistance due to the air (drag) decreases also. This would be relevant if you were traveling say to the Rockies to skydive or any place with much higher elevations. Since the air resistance is less at higher altitudes, your descent rate under canopy and loss of altitude while doing 360s would increase.

Hope that helps

There is a lot more to know and understand about physics and skydiving. In fact I am writing a term paper this semester which I will put on DZ.com since I promised a few people I would let them read it. They gave me some pretty good resources.

Roy Bacon: "Elvises, light your fires."

Sting: "Be yourself no matter what they say."

[NoShitThereIWas 0]

Sean, I think I understand what you are saying. If there was no air resistance (i.e. you were in a vacuum) two objects with different weights and or shape would hit the ground at exactly the same time because gravity is acting on them equally, pulling them both toward the earth at -9.8 m/s^2. However, when you factor in the air resistance which is the result of molecular air collisions on a moving object (in this case the leading surface of a skydiver's body) a change in weight, speed and/or surface area will affect the time it takes for a jumper to reach his/her terminal velocity. These factors are also how we modify our fall rate.

Roy Bacon: "Elvises, light your fires."

Sting: "Be yourself no matter what they say."

[kizza 0]

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dropping two objects one extremely heavy and one that is light and both hitting the ground at the same time

Remember, newtons law of gravity states that all objects accelerate towards earth at the same rate, that is 9.8m/s.

The law doesn't take into account terminal velocity, and you didn't have to take it into account either, because your objects never got fast enough to reach it.

However at 14,000 feet we can quite easily get up to terminal velocity, and it is reached when our weight is equal to our drag, or air resistance force. Where we can't fall any faster (unless we change body position). If we are heavier it will take more air resistance to counteract that weight. Hence, a heavier person will accelerate to a higher terminal velocity than a light person, and fall at a different rate.

http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/newtlaws/sd.gif

This diagram illustrates terminal velocity, but for a parachute aswell. Essentially it's the same thing. We have a terminal velocity for freefall, and a slower terminal velocity under canopy. Under canopy we fall much slower than in freefall due to the huge surface area we use to create air resistance. Because of that huge surface area, the air resistance required to counteract our weight can be acheived at a much lower speed, which means we can land safely :)

This brings me to a question that has always puzzled me: How do people know how to make those space balls fall at the same rate as them? How heavy are they?

[nacmacfeegle 0]

Okay Sean,I'm not get all techno on ya, everyone else has done an admirable job of that already. Sometimes its best not to think abut why, just focus on the how.....

Ways to speed up,
wear weight, get a more 'slick' suit material, change your body position, or change your attitude (oreintation) to the wind.
Ways to go slower
Lose weight, get a 'rougher' suit, get a suit that dramatically changes your body shape, like a wingsuit, a baggy suit, or a birdman suit, or change your body position or attitude (orientation) to the wind.

Some of these you can do to minor effect during a skydive, (changing body position) others may have to be considered when you are gearing up for dive (wearing weight or selecting the right suit). All are useful and have a part to play in skydiving.

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He who receives an idea from me, receives instruction himself without lessening mine; as he who lights his taper at mine, receives light without darkening me. Thomas Jefferson

[kallend 1,661]

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I am new to the sport and I am currently reading the book "skydivers handbook" In this book the author states that Weight and Drag are both Ratios that determine speed. I am confused I remember back in middleschool doing experiments while dropping two objects one extremely heavy and one that is light and both hitting the ground at the same time. Newtons law of gravity maybe... anyways it seems like that would apply towards skydiving as well which means weight is not a factor but simply the amount of surface area that the diver is. Anybody wanna clue me in? Thx

Sean

Check out www.iit.edu/~kallend/skydive/ for a Powerpoint presentation on Physics of Skydiving, aimed at high school level. It's meant to be used along with clips from Point Break and similar movies.

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

[kallend 1,661]

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>At low speeds, air resistance is almost proportional to speed . . .

At low speeds, power required to move a _car_ is proportional to speed, but that's because rolling resistance dominates. I don't think air resistance scales that way.

It does if the Reynolds number is < 1 (not that this has any relevance to skydiving).

...

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