Fall rate

[loch1957 0]

Just a question,is there a formula for figuring landing speed based on increase or decreased weight on same canopy. For instance if I gain ten how much more speed does that weight cause?.

Avoiding danger is no safer in the long run than outright exposure. Life is either a daring adventure, or nothing.”

[Deisel 37]

The term fall rate is used to describe terminal velocity in freefall. What you are asking about has to do with wing loading. Each model canopy has a different descent rate that is determined by how it's made. The jumper affects this by how much weight is suspended under the canopy. More weight = faster canopy (if all other inputs are equal).

The formula used is exit weight divided by canopy square footage. For example; 220lbs exit weight (jumper + all equipment) on a 190sq ft canopy equals a wing load of 1.15:1 ratio.

And for a comparison, we typically start student jumpers with a very conservative 0.7:1 ratio in order to slow them down. A competition swooper could easily be in the 2.5:1 range, which is extremly aggressive.

D

The brave may not live forever, but the timid never live at all.

[pchapman 262]

We usually say 'descent rate' when talking about the canopy flight rather than freefall.

The basic aeronautical equations show that, all else being equal, the airspeed in a glide will vary with the square root of the wing loading.

That will be sufficiently accurate for our purposes when you change the wing loading by changing the weight. If changing the wing loading by changing the canopy size, the formula isn't quite as good but is still a good approximation. (Other factors come into play even for the same type of canopy: You can change the size of the canopy 20% but you don't change the pilot chute, lines, or your own body by the same 20% in area. Not everything scales up or down the same amount.)

So:
Double the wing load = 41% more speed (root 2 = ~ 1.41)

Or for example, adding 20% to the wing loading is a 1.2 ratio, take the square root = 1.095, so the increase in flight speed will be 9.5%. This applies both to forwards speed and downward speed. Glide ratio is unaffected by wing loading, at the first level of approximation.

How the actual flare on the canopy works will depend on other factors (e.g., a given canopy can get appreciably harder to land beyond some high wing loading). Things like wind and human limits can also affect the perception of landing. (E.g., 2 mph more might not mean much unless it goes from a speed you personally can easily run out to a speed you can't easily run out.)

[CygnusX-1 42]

That is all good information - however useless to the OP.

To rephrase the question and use your data:

Given a wing loading of 1.15:1, what is the forward speed increase when the skydiver increases his/her wing loading to 1.21:1?

This is on the same canopy, same wind conditions, same altitude (MSL), same approach style, maybe 30 minutes later in time, same square footage of surface area of skydiver, same suit, same helmet, same lines on the canopy, same style collapsing the slider, same pilot chute, exact same location of loosened chest strap, etc, etc, etc.

Oh ya, I have no clue so I can't answer it.

[Deisel 37]

Thanks a lot, smart ass!

Too funny.

The brave may not live forever, but the timid never live at all.

[erdnarob 1]

I know only two persons who have studied scientifically low speed aerodynamics concerning parachutes and who can help you for that. They are John Sherman from Jump Shack and Jean Potvin from Parks College of St Louis University. Both of them have given seminars about this subject at the Parachute Industry Association (PIA) symposium.

What I can tell you is the following : In aerodynamics there is a well known formula : F=KSV^2, where F is the force generated by a fluid (air) flowing against the surface of a body, K is a constant depending on the density and viscosity of the fluid, the units you use and the shape of the body (CX), S is the surface or area opposed to the fluid flow and V is the relative speed of the fluid with respect to the area.

Note 1: V is to the square which means for instance that if the surface of the body opposed to the fluid is multiplied by 2, the speed will be divided by the square root of 2 (this when the force stay the same (your weight in free fall for instance))

Note 2 : the formula above is more suitable for round parachutes or somebody in free fall. For square parachutes, they have more or less the shape of an airfoil and have to be studied accordingly. There are 4 forces on an airfoil : 1) the trust (given by a gravity component on a parachute) 2) the drag (or friction thru the air) 3) the lift (generated by Bernoulli's force), 4) the load which is the suspended weight

Learn from others mistakes, you will never live long enough to make them all.

[erdnarob 1]

Sorry... if you gain 10 pounds...I just forgot.

According the formula F=KSV^2 :

10 pounds increase from the original weight of (for exemple) 180 pounds which makes 190 pounds.

This is an increase by 10/180 = 0.05555 or 5.55% of the original weight. That means your original weight has been multiplied by 1.0555. Therefore your speed would be multiplied by the square root of 1.0555 which is 1.0274 equivalent to an speed increase of 2.74% (not too much). But as I said this is true for round parachute but can give you an idea for the landing speed of a square parachute.
Note : when the load on a square parachute is increased the glide ratio (distance vs altitude) stays about the same which means both the forward and vertical speeds are increased with same proportions.

Learn from others mistakes, you will never live long enough to make them all.

[Southern_Man 0]

pchapman, this is very helpful, thank you.

"What if there were no hypothetical questions?"

[loch1957 0]

Thanks guys or gals, i can figure it out from this.

Avoiding danger is no safer in the long run than outright exposure. Life is either a daring adventure, or nothing.”

[potatoman 0]

The answer that is 100% correct is 2.5%, roughly. hahahaha - Too many variables

You have the right to your opinion, and I have the right to tell you how Fu***** stupid it is.
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Whatever you do, don't listen to ChrisD.

[kallend 1,635]

Quote

Sorry... if you gain 10 pounds...I just forgot.

According the formula F=KSV^2 :

10 pounds increase from the original weight of (for exemple) 180 pounds which makes 190 pounds.

This is an increase by 10/180 = 0.05555 or 5.55% of the original weight. That means your original weight has been multiplied by 1.0555. Therefore your speed would be multiplied by the square root of 1.0555 which is 1.0274 equivalent to an speed increase of 2.74% (not too much). But as I said this is true for round parachute but can give you an idea for the landing speed of a square parachute.
Note : when the load on a square parachute is increased the glide ratio (distance vs altitude) stays about the same which means both the forward and vertical speeds are increased with same proportions.

It would be reasonable for a square parachute too, since the formula for lift varies with speed in exactly the same way as the formula for drag.

...

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