Integrating a squared velocity

In summary, the conversation discusses how to integrate an equation in order to obtain velocity and time of flight for a ballistics calculator. The equation is rearranged and integrated, taking into account initial conditions. The conversation also addresses issues with units and accuracy in the solution.
  • #1
gendoikari87
11
0
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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  • #2
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.
 
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  • #3
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
 
  • #4
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?
 
  • #5
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.
 
  • #6
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
 
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  • #7
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).
 
  • #8
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
 
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  • #9
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.
 
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  • #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 dimention 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 dimention 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.
 
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  • #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.
 

1. What is the equation for integrating a squared velocity?

The equation for integrating a squared velocity is ∫ v^2 dt = (v^3)/3 + C, where v is the velocity and C is the constant of integration.

2. Why is integrating a squared velocity useful in science?

Integrating a squared velocity is useful in science because it allows us to determine the distance an object has traveled over a given time period. This is important in many areas of science, such as physics and engineering.

3. What is the physical interpretation of integrating a squared velocity?

The physical interpretation of integrating a squared velocity is that it represents the change in an object's kinetic energy over a given time period. This is because kinetic energy is equal to 1/2 mv^2, where m is the mass of the object and v is the velocity.

4. Can you integrate a squared velocity if the velocity is not constant?

Yes, you can still integrate a squared velocity if the velocity is not constant. However, this will result in a more complex equation, as the velocity will need to be represented as a function of time in order to be integrated.

5. How is integrating a squared velocity related to finding the area under a curve?

Integrating a squared velocity is related to finding the area under a curve because the integral of v^2 is equal to the area under the curve of v^2. This is a useful concept in calculus and can be applied to various real-world scenarios in science and engineering.

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