Solve DiffEq for Runge-Kutta 4th Order Input

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Discussion Overview

The discussion revolves around converting differential equations into a format suitable for input into a Runge-Kutta 4th order spreadsheet. Participants explore how to express velocity derivatives in the form y' = ... and address the complexities of a three-dimensional model of ball flight.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant seeks assistance in converting the equation dvx/dt = -FDvx/mv into the form y' = ...
  • Another participant suggests defining y = vx to achieve the required form.
  • A participant proposes that defining y = ax leads to dy/dt = -FDy/mv, questioning if this is correct.
  • There is a suggestion that the expression should include v defined as sqrt(vx2 + vy2), indicating that not all variables and equations were listed.
  • A participant clarifies that v should be defined as sqrt(vx2 + vy2 + vz2) for a three-dimensional model, raising questions about handling additional derivatives (vy & vz).
  • Another participant indicates that in this case, y would be treated as a vector (y1, y2, y3), leading to three equations for the system.

Areas of Agreement / Disagreement

Participants generally agree on the approach to defining variables and converting the equations, but there are uncertainties regarding the treatment of additional derivatives and the completeness of the equations presented.

Contextual Notes

There are limitations regarding the completeness of the equations and the definitions of variables, as not all necessary information has been provided by the original poster.

andyfive
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Hi,

Hi. I am trying to enter some differential equations into a Runge-Kutta 4th order spreadsheet which requires input in the form y' = ...? My differential equations are in the form :
dvx/dt = -FDvx/mv (i.e relate velocity derivative with respect to time to acceleration).
Can anyone please help or point me in the right direction as to how to convert this to to y' = ...? format?

Many Thanks
 
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Hi andyfive, welcome to PF! :smile:

Perhaps if you define y = vx you have the required form?

Or possibly y = (vx, vy)?
 
Many Thanks for your reply I like Serena. If I understand correctly, by defining y = ax I get :

dy/dt = -FDy/mv

If y' is equal to dy/dt then it is as simple as saying y' = -FDy/mv. Is this correct?

Thanks
 
andyfive said:
Many Thanks for your reply I like Serena. If I understand correctly, by defining y = ax I get :

dy/dt = -FDy/mv

If y' is equal to dy/dt then it is as simple as saying y' = -FDy/mv. Is this correct?

Thanks

Basically, yes!

However, you also have a "v" in your expression.
I suspect this should actually be sqrt(vx2+vy2), but I'm guessing here, since I think you did not list all your variables and equations.
If it is, "y" would come back into the expression for "v".
 
I like Serena said:
Basically, yes!

However, you also have a "v" in your expression.
I suspect this should actually be sqrt(vx2+vy2), but I'm guessing here, since I think you did not list all your variables and equations.
If it is, "y" would come back into the expression for "v".

Thanks again I like Serena. Yes, you are absolutely correct, the v will need to be defined actually as sqrt(vx2+vy2+vz2) - this is a model for a 3 dimensional ball flight. The initial velocity(v) at time(t) = 0 is known and hence the three initial velocity vectors are known. Not sure what happens though when other derivatives are part of the expression (ie vy & vz)?

Regards
 
andyfive said:
Thanks again I like Serena. Yes, you are absolutely correct, the v will need to be defined actually as sqrt(vx2+vy2+vz2) - this is a model for a 3 dimensional ball flight. The initial velocity(v) at time(t) = 0 is known and hence the three initial velocity vectors are known. Not sure what happens though when other derivatives are part of the expression (ie vy & vz)?

Regards

In this case you would have y = (y1, y2, y3) as a vector.

And you would effectively have 3 equations, the first being:

\frac {dy_1} {dt} = \frac {-F_D y_1} {m \sqrt{y_1^2+y_2^2+y_3^2}}
 
I like Serena said:
In this case you would have y = (y1, y2, y3) as a vector.

And you would effectively have 3 equations, the first being:

\frac {dy_1} {dt} = \frac {-F_D y_1} {m \sqrt{y_1^2+y_2^2+y_3^2}}

Brilliant. Many Thanks I like Serena. You really have helped clarify things for me. I'll get back to the spreadsheet and see what results I now obtain.
 
andyfive said:
Brilliant. Many Thanks I like Serena. You really have helped clarify things for me. I'll get back to the spreadsheet and see what results I now obtain.

Cheers! :smile:
 

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