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

• Can.i.say?
In summary, the speaker is a skydiving coach who is trying to determine the force of wind moving at approximately 120 mph on an object with a surface area of 25 square inches. They mention their experience in a vertical wind tunnel and their interest in knowing how slow they can fall. They also mention that their body position and surface area plays a role in their fall speed. The speaker is looking for a general formula to calculate this force and mentions the drag equation as a possible solution. They also discuss the drag coefficient and how it affects their fall speed.
Can.i.say?
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

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

A.T. said:
Is there something more simple? Area, wind speed=lbs?

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?

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.

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.

zoki85 said:
He wants to know how slow he can fall.
No, he already knows it's 90 mph.

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...

## 1. How do you calculate the force of wind on a flat object?

The force of wind on a flat object can be calculated using the formula F = 0.5 x ρ x A x V2, where F is the force in Newtons, ρ is the density of air in kg/m3, A is the surface area of the object in square meters, and V is the wind speed in meters per second.

## 2. Why is it important to calculate the force of wind on a flat object?

Calculating the force of wind on a flat object is important for safety reasons, especially in activities like skydiving. It helps determine the maximum wind speed that an object can withstand without causing damage or danger to the people involved.

## 3. What factors affect the force of wind on a flat object?

The force of wind on a flat object is affected by the density of air, surface area of the object, and wind speed. Other factors such as the shape and orientation of the object, as well as air resistance and turbulence, can also play a role in the calculation.

## 4. How does air density affect the force of wind on a flat object?

The denser the air, the greater the force of wind on a flat object. This is because denser air has a higher mass per unit volume, resulting in a greater impact on the object. Altitude, temperature, and humidity are all factors that can affect air density.

## 5. Can the force of wind on a flat object be reduced or eliminated?

The force of wind on a flat object cannot be completely eliminated, but it can be reduced by changing the orientation or shape of the object, or by using materials with lower air resistance. However, in extreme wind conditions, it may not be possible to completely eliminate the force of wind on a flat object.

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