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gendoikari87
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Okay I'm working on making a ballistics calculator and I need to know how to integrate this.
To get velocity, and ultimately to get time of flight. So that I can use that to determine drop with an angle of 0.
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).nasu said:You will have dv/dT=-kv^2. This can be solved by direct integration, after rearranging a little bit.
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
rcgldr said:Don't forget to include a constant after the integration.
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.gendoikari87 said:... that's just initial velocity is it not?
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
No. Look at the units.gendoikari87 said:So this gives me:
1/kv+vi=T correct?
Why doesn't seem right?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,
On the other hand, this does not seem right at all.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
nasu said:Why doesn't seem right?
Of course, you should add the constant that takes into account the initial position.
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.
This may happen very well because you have only the drag force and no other forces.gendoikari87 said:Problem is the solution with the log function is giving me ansers of several thousand years for a flight time
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).
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.
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.
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.
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.
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.