Calculating Force of Wind on Flat Object: A Skydiving Coach's Guide

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SUMMARY

The discussion focuses on calculating the force of wind on a flat object with a surface area of 25 square inches at a wind speed of 120 mph. The drag force can be calculated using the formula F_d = 0.00256 * k * V^2 * (A/144), where V is the wind speed in mph, A is the area in square feet, and k is the shape factor. For a flat object, the drag force at 90 mph results in approximately 5 pounds of air drag force on each hand when extended. The key takeaway is that increasing the exposed surface area reduces the falling speed due to increased drag.

PREREQUISITES
  • Understanding of the drag equation and its components
  • Familiarity with wind speed measurements in mph
  • Basic knowledge of physics related to forces and motion
  • Awareness of drag coefficients for different shapes
NEXT STEPS
  • Research the drag coefficient values for various body positions in skydiving
  • Learn how to apply the drag equation to different surface areas and shapes
  • Explore the effects of turbulence and compression on drag force
  • Investigate terminal velocity calculations for different skydiving positions
USEFUL FOR

Skydiving coaches, physics enthusiasts, and anyone interested in understanding the dynamics of air resistance and drag forces in skydiving scenarios.

Can.i.say?
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I am a skydiving coach and I am trying to figure out how many pounds of force wind moving approximately 120 mph has on something with a surface area of 25 square inches. I know turbulence and compression play into it as we'll but I just want a general formula to figure it out. The object is flat. I have spent over 500 hours in a vertical wind tunnel teaching people how to manipulate the wind to fly their bodies. I want to know how slow I can fall
 
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120 mph is roughly the speed of the wind for a skydiver, right? So what is the average person's cross sectional area...?
 
russ_watters said:
120 mph is roughly the speed of the wind for a skydiver, right? So what is the average person's cross sectional area...?
What do you mean cross sectional area?
 
russ_watters said:
120 mph is roughly the speed of the wind for a skydiver, right? So what is the average person's cross sectional area...?
I know I have been clocked at 210 mph fastest and 90 mph approx. slowest. I am not sure of average area of someone skydiving on with their stomach to the wind. But I was more thinking about the lift I get from my hands. Approx. 25 square inches
 
I'm not a skydiver, but you fall slower at terminal velocity when you expose as much of your body (including hands) to the 'wind'. Belly down is slower than feet first, because you expose greater surface area to the air, which increases the drag. At terminal velocity, the skydivers weight is equal to the air drag force. Without getting into the actual Physics too much, the drag force at terminal velocity V on a cylindrical surface is 0.00256V^2( A) , where V is in mph and A is the area of the body exposed to the wind, in square feet. So if a person with gear weighs 200 pounds and exposes full body to the wind (say 10 square feet to use a number) the air drag force at terminal velocity is 200 pounds, solve for V and you get about 90 mph. If you fall feet first if that is possible, then your exposed surface area is about say I don't know about 2 square feet, so V works out to about 200 mph . These are real rough numbers, depends on shape factor whether surface area is flat or cylindrical or rolled up a ball, and other factors.
 
A.T. said:
Area & wind speed is all you need there. You can look up the drag coefficient in some tables:
http://en.wikipedia.org/wiki/Drag_coefficient
And which one of these profiles is his body in a horizontal position?
 
  • #10
zoki85 said:
And which one of these profiles is his body in a horizontal position?
He didn't ask about his body, but something that is flat and has a surface area of 25 square inches.
 
  • #11
A.T. said:
He didn't ask about his body, but something that is flat and has a surface area of 25 square inches.
He wants to know how slow he can fall. In the wiki page you linked to, there's actually listed drag coefficient range for a "man in upright position"
(= 1-1.3). If it is measured, numerically calculated or somehow aproximated I don't know, but maybe could be of use.
 
  • #12
zoki85 said:
He wants to know how slow he can fall.
No, he already knows it's 90 mph.
 
  • #13
Can.i.say? said:
I know I have been clocked at 210 mph fastest and 90 mph approx. slowest. I am not sure of average area of someone skydiving on with their stomach to the wind. But I was more thinking about the lift I get from my hands. Approx. 25 square inches
At 90 mph, the drag force on each outstretched flat- to -the- wind 25 square inch hand is 0.00256(k)(90)^2(25/144), where k is the shape factor (I'll use k= 1.3), which works out to about a mere 5 pounds of air drag force on each hand. So when falling straight down belly down with hands closed tight to your body versus hands open and extended away from your body, you increase your exposed area from the assumed 10 square feet to perhaps let's say 11 square feet including extended arms , and now your speed instead of being 90 mph is more like 85 mph, or that order of magnitude. As you know, the more surface area of your body you can expose to the wind, the slower will you fall. And you want a large drag factor...'flat out' versus 'curled up' yields a greater drag...
 

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