Projectile Motion with Air Resistance Excel EVIL

[gMitchell]

Homework Statement

Ok I am working on a spreadsheet that calculates the velocity/position/drag of a point-mass projectile and I have been digging up the same equations for days and they aren't helping... lol... I just can't seem to get the numbers right, any help would be greatly appreciated :)

Homework Equations

Vx = Vx(0) + t*Ad,x(t)
Vy = Vy(0) + -9.8*t + t*Ad,y(t)
...
Fd = Drag Coefficient * Air Density * Velocity^2 (how can drag be dependent on velocity and vice versa? This causes a circular reference problem in Excel...) * Frontal Area

The Attempt at a Solution

Gah... more possibilities than exist in the seventh dimension... lol

 

[fatra2]

Hi there,

It has been a very long time, but your equation seem a bit ackward. If you are trying to evaluate the velocity and position of a projectile, you need to have a complete equation that describes it in 2D. For that you need to consider a partial differential equation, since the drag force will depend on the velocity of the object.

Cheers

 

[gMitchell]

:) thanks for the response... I am a little rusty with my calculus lol, could you explain to me how to do it that way? How would I be able to incorporate this into a spreadsheet? Thanks :)

 

[fatra2]

Holly cow, this is far back, but here goes nothing.

The acceleration of the object is:
\vec{a}=\frac{d^2\vec{r}}{dt^2}
Very simple up to now. Thinking of what is the object acceleration, you can imagine that the drag makes the object slow down, and gravity is also present. Therefore:
\vec{a} = \lambda \vec{v} + \vec{g}
Where \lambda is the drag coefficient.

With the xy notation:
a_x = -\lambda v_x = \frac{d^2x}{dt^2}
a_y = - \lambda v_y - g = \frac{d^2y}{dt^2}

At last:
-\lambda \frac{dx}{dt} = \frac{d^2x}{dt^2}
-\lambda \frac{dy}{dt} - g = \frac{d^2y}{dt^2}

You just need to solve these two equation and you will get the xy position of your object throughout the trajectory.

Cheers

 

[gMitchell]

Thank you so much! This is much easier to understand than what google and wikipedia brought up. :) I really appreciate this! Amazingness!

 

[badphysicist]

Also, if you are solving numerically (like in Excel for instance) you need to discretize the equations. This basically means that d's in the dx's and dt's (and so forth) become delta's (finite steps). The smaller the step, the more accurate your solution is, but the more steps you have to use to get to your end point.

 

[fatra2]

Hi there,

Also, if you are solving numerically (like in Excel for instance) you need to discretize the equations. This basically means that d's in the dx's and dt's (and so forth) become delta's (finite steps). The smaller the step, the more accurate your solution is, but the more steps you have to use to get to your end point.

You are absolutely right. This would be solving the equations numerically. This would give you a solution that approximate the real solution.

What I was thinking, at first, was that you can solve these two equations. Let's face it they are not the world to solve, and just plot the solution into an excel graph!

Cheers

 

[mysticjbyrd]

When I solve,
<br /> -\lambda \frac{dx}{dt} = \frac{d^2x}{dt^2}<br />

x(t) = c_2-(c_1 e^(-t Z))/Z