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Mathmatically how would i account for it?

Say after one second of free fall how much of my velocity would be lost due to air?

Im assuming it has to do with the ratio of mass of the object to its crossection to the air

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- Thread starter jonatron5
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- #1

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Mathmatically how would i account for it?

Say after one second of free fall how much of my velocity would be lost due to air?

Im assuming it has to do with the ratio of mass of the object to its crossection to the air

- #2

Buzz Bloom

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Hi jonatron:

I suggest you use a browser and search on "air resistance formula".

Regards,

Buzz

I suggest you use a browser and search on "air resistance formula".

Regards,

Buzz

- #3

FactChecker

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If the fall is short, you can assume that the air density is a constant. In general, the drag force must be calculated frequently for the changing air density and any rotation of the object that presents a different cross section. Then the motion of the object can be calculated for that force till the next drag calculation is done.

- #4

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Drag does not care about the mass of the object, only the shape and surface.

The drag force itself does not, but its effect on the motion of the object certainly depends on the object's mass given that ##a = F/m##.

It is proportional to velocity squared, air density, the cross sectional area of the object and a coefficient of drag that represents the shape of the object. See the first equation in https://en.wikipedia.org/wiki/Drag_(physics).

I'd also like to point out (for the sake of the OP) that drag coefficients can be incredibly useful, but are ultimately just empirical approximations. Drag is actually an extraordinarily complicated quantity that, for many shapes (e.g. a sphere), can be estimated quite readily and accurately with a drag coefficient. More complex shapes (e.g. an airplane) are more difficult to handle.

If the fall is short, you can assume that the air density is a constant. In general, the drag force must be calculated frequently for the changing air density and any rotation of the object that presents a different cross section. Then the motion of the object can be calculated for that force till the next drag calculation is done.

Alternatively, just set the whole thing up as a differential equation and solve it to get a complete time history of the position, velocity, and drag for a given object. This can often be done without too many unreasonable assumptions for simple situations like a falling ball.

- #5

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$$f=f(v)\hat { v }$$

The function f(v) varies with v in a very complicated way. However, at lower speed(lower than the speed of sound), it is a good approximation to write

$$f(v)=bv+c{ v }^{ 2 }$$

The linear term arises from the viscous drag of the medium and is generally proportional to the viscosity of the medium and the linear size of the object. The quadratic term is proportional to the density of the medium and the cross-sectional area of the object.

For big and fast objects, the quadratic resistance dominates, for small and slow objects, linear resistance dominates.

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