Integrating a squared velocity

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    Integrating Velocity
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Discussion Overview

The discussion revolves around the integration of a squared velocity in the context of developing a ballistics calculator. Participants explore the mathematical formulation necessary to derive time of flight and distance traveled under the influence of drag, particularly focusing on the integration of velocity equations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks to integrate velocity to determine time of flight and drop for a projectile.
  • Another suggests rearranging the equation to express it in terms of velocity, leading to the form dv/dT = -kv^2, which can be integrated directly.
  • Several participants discuss the need to include a constant after integration and how to apply initial conditions to solve for it.
  • There is a proposal that the integration leads to a function of velocity versus time, and the initial velocity must be accounted for in the integration process.
  • One participant expresses confusion over the resulting equations and questions the validity of their integration steps.
  • Another participant points out potential issues with mixing variables during integration, suggesting a need for separation before proceeding.
  • Concerns are raised about the physical implications of the solutions, particularly regarding the time of flight calculations yielding unrealistic results.
  • Participants discuss the significance of drag force and its impact on the projectile's motion, noting that neglecting other forces can lead to exaggerated flight times.

Areas of Agreement / Disagreement

Participants express various viewpoints on the integration process and the resulting equations, with no consensus reached on the correctness of the approaches or the final forms of the equations. Disagreements arise regarding the handling of constants and the implications of the drag coefficient.

Contextual Notes

Participants note potential limitations in their approaches, including unresolved assumptions about the drag coefficient and the need for unit analysis. The discussion highlights the complexity of integrating velocity equations under the influence of drag, with varying interpretations of the mathematical steps involved.

gendoikari87
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Okay I'm working on making a ballistics calculator and I need to know how to integrate this.

gif.latex?\frac{d^{2}X}{dT^{2}}=-k(\frac{dX}{dT})^{2}.gif
To get velocity, and ultimately to get time of flight. So that I can use that to determine drop with an angle of 0.
 
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Try to solve the equation in v first.
Replace dX/DT by v (on both sides).
You will have dv/dT=-kv^2. This can be solved by direct integration, after rearranging a little bit.
 
Last edited:
nasu said:
You will have dv/dT=-kv^2. This can be solved by direct integration, after rearranging a little bit.
Rearraging, this is what you would integrate, don't forget to include a constant after the integration (then solve for that constant based on initial conditions).

dv / (-k v2) = dt
 
rcgldr said:
Rearraging, this is what you would integrate, don't forget to include a constant after the integration (then solve for that constant based on initial conditions).

dv / (-k v2) = dt

... that's just initial velocity is it not?
 
rcgldr said:
Don't forget to include a constant after the integration.

gendoikari87 said:
... that's just initial velocity is it not?
The integral will produce a function of velocity versus time, so the constant would be added or subtracted from the function of velocity at time = zero to account for the initial velocity.
 
rcgldr said:
Rearraging, this is what you would integrate, don't forget to include a constant after the integration (then solve for that constant based on initial conditions).

dv / (-k v2) = dt

So this gives me:

1/kv+vi=T correct?

From here I changed the V to dx/dt but since it is 1/v it becomes

(1/k)*dt/dx+Vi=T

Multiplying by dx I then get

(1/k)dt+Vidx=Tdx

Integrating yields

T/k +ViX=T*X (accounting initial time and distance as zero)

But that just seems wrong...

Breaking it down all the way I get :

(Vi*X)/(x-1/k)=T
 
Last edited:
gendoikari87 said:
So this gives me:

1/kv+vi=T correct?
No. Look at the units.
You add initial speed and inverse speed (times k).
It would be more obvious if you put it like
dv/v^2=-kdt
The integration on both sides gives
-1/v=-kt +C

At t=0 v=vi so
-1/vi=C or C=-1/vi
Then you have
1/v=1/vi+kt
Solve this for v and you have v(t).
Then you can integrate v(t) over time to get x(t).
 
Cool, so I've solved for V:

V= 1/(KT+1/Vi)

Wolfram integrates it as

Log(KTVi+1)/K=X

That doesn't seem right,

If I simply integrate KT+1/Vi with respect to X (because 1/v is dT/dX) I get

KTX+X/VI=T

Which becomes

X/(Vi-KXVi)=T
 
Last edited:
gendoikari87 said:
Cool, so I've solved for V:

V= 1/(KT+1/Vi)

Wolfram integrates it as

Log(KTVi+1)/K=X

That doesn't seem right,
Why doesn't seem right?
Of course, you should add the constant that takes into account the initial position.

gendoikari87 said:
If I simply integrate KT+1/Vi with respect to X (because 1/v is dT/dX) I get

KTX+X/VI=T

Which becomes

X/(Vi-KXVi)=T
On the other hand, this does not seem right at all.
If you write
1/v=dt/dx=kt+1/vi
you have mixed variables (t and x). You cannot integrate right away.
After separation you have
dx=dt/(kt+1/vi) which seems to be same as above.
 
Last edited:
  • #10
nasu said:
Why doesn't seem right?
Of course, you should add the constant that takes into account the initial position.

call it intuition. I'll have to do unit analysis though. K I think has a dimension but IIRC it's inverse mass I'll check after while.
 
  • #11
Sorry, I was editing my post while you posted your reply.
See above.
 
  • #12
gendoikari87 said:
call it intuition. I'll have to do unit analysis though. K I think has a dimension but IIRC it's inverse mass I'll check after while.

The dimension of k is inverse length.And kt is inverse speed.
 
  • #13
Problem is the solution with the log function is giving me ansers of several thousand years for a flight time

28560fb9ccae7b5f811de11f965d5478.png
This is where I get K, it's everything but the V squared term and divided by mass.

Edit: hold on, it could be my drag co-efficient is f'ed up.
 
Last edited:
  • #14
Yup, it was the BC, or at least i think, I'll have to check it out some more. Thanks for the help.
 
  • #15
gendoikari87 said:
Problem is the solution with the log function is giving me ansers of several thousand years for a flight time
This may happen very well because you have only the drag force and no other forces.
After some time (and distance traveled) the speed will decrease practically to zero.
If your distance is much larger than the distance at which speed becomes very small, the time becomes very large. A plot of t(x) will show you better this behavior.
For example, taking vi=10 m/s and k=1m^-1, the time to travel 2 m is 0.6 s but the time to travel 20 m is of the order of 10^7 s (about a year).
 
  • #16
nasu said:
This may happen very well because you have only the drag force and no other forces.
After some time (and distance traveled) the speed will decrease practically to zero.
If your distance is much larger than the distance at which speed becomes very small, the time becomes very large. A plot of t(x) will show you better this behavior.
For example, taking vi=10 m/s and k=1m^-1, the time to travel 2 m is 0.6 s but the time to travel 20 m is of the order of 10^7 s (about a year).

Nah I was taking it at 100 meters. It was unit problem with the Drag coefficient converting from BC.
 

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