Air Drag Formula: Solving Maik's Rocket Project Challenge

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SUMMARY

This discussion centers on the challenges faced by Maik in calculating the height of a solid fuel rocket while accounting for air drag. The air drag force is defined by the equation Fd(t) = -0.5 * p * A * Cd * v^2, which complicates the calculation as both the rocket's speed and acceleration depend on air drag. The conversation highlights the necessity of using differential equations to solve for air drag, as the drag force is inherently linked to the rocket's changing velocity. Additionally, it emphasizes that no single formula effectively models drag at high speeds, particularly in transonic and supersonic regimes.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with differential equations
  • Knowledge of fluid dynamics principles, particularly air drag
  • Basic calculus, specifically integration and differentiation
NEXT STEPS
  • Research methods for solving differential equations in physics
  • Explore computational fluid dynamics (CFD) tools for simulating air drag
  • Study the effects of varying drag coefficients (Cd) at different speeds
  • Learn about numerical methods for approximating solutions to complex equations
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Aerospace engineers, physics students, and hobbyists involved in rocketry or fluid dynamics who seek to understand the impact of air drag on rocket performance.

v6maik
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Hello,

I'm working on a project about solid fuel rockets and since these are moving through the atmosphere, they experience Air-drag. I'm trying to set up a formula to exactly calculate the height a rocket will achieve. So without the use of any model. The problem I ran into is that I can't get the formula for acceleration to include air-drag, since I ran into the following loop:

-the speed of the rocket (and thereby the acceleration) depends on the air-drag.
-the air-drag depends on the speed of the rocket (and thereby the acceleration)

I found this equation for air-drag but it's of no use in its current form.

Fd(t)= -0,5 * p * A * Cd * v^2

as you can see, drag depends on the speed relative to the air-mass, which is pretty obvious.

But since v=a * t, this formula is the same as:

Fd(t)= -0,5 * p * A * Cd * (a*t)^2

and since a= F/m this formula is the same as:

Fd(t)= -0,5 * p * A * Cd (((F / m ) *t)^2

note that F is the net Force on the rocket. The net Force at a given time is equivalent to the propulsion force minus gravity minus drag:

Fnet(t)= Fp(t) - Fg(t) - Fd(t)

So the drag formula now is:

Fd(t)= -0,5 * p * A * Cd (((Fm(t) - Fg(t) - Fd(t)) / m ) *t)^2

Notice that this formula Fd(t) involves it's own answer, so it is a differential equation, right?

Now, I can simplify this formula to this, leaving 3 constants: a, b and c:

Fd(t)= a * ( (b-Fd(t) )/c * t)^2

Any suggestions about solving this problem? Or might there be a different equation to calculate air-drag at a given time during acceleration?

Thanks ahead!

Kind regards,
Maik
 
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There isn't anyone formula for drag that works for high speeds, especially transonic and supersonic speed. The mathematical model for bullets and cannon shells is complex, relies on the usage of tables from actual measured data, and it wasn't until around 1990 (Desert Storm), that tanks (USA tanks) could reliably shoot other tanks with a single shot with advanced fire control systems.
 
v6maik said:
Hello,

I'm working on a project about solid fuel rockets and since these are moving through the atmosphere, they experience Air-drag. I'm trying to set up a formula to exactly calculate the height a rocket will achieve. So without the use of any model. The problem I ran into is that I can't get the formula for acceleration to include air-drag, since I ran into the following loop:

-the speed of the rocket (and thereby the acceleration) depends on the air-drag.
-the air-drag depends on the speed of the rocket (and thereby the acceleration)

I found this equation for air-drag but it's of no use in its current form.

Fd(t)= -0,5 * p * A * Cd * v^2

as you can see, drag depends on the speed relative to the air-mass, which is pretty obvious.

But since v=a * t,

no. that is only true if the acceleration is constant (and if the velocity is zero at t=0). In general
<br /> a=\frac{dv}{dt}<br />
and so
<br /> v(t)=v(0)+\int_0^t a(t) dt<br />
 

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