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#Actualrockpool

Posted 17 April 2013 - 10:10 AM

Hi,

I've got an example from a book working, however the formulas are so densely packed i'm having trouble really understanding it.

What i'm after is extracting the relevant information to extend the example and implementing a bounce.

The book is here, and the sample code can be found on the right, i'm using the Cannon 2 example:

http://shop.oreilly.com/product/9780596000066.do

 

Here's the DoSimulation function (it includes the preliminaries in each step for clarity)

    double    cosX;
    double    cosY;
    double    cosZ;
    double    xe, ze;
    double    b, Lx, Ly, Lz;
    double    tx1, tx2, ty1, ty2, tz1, tz2;

    // new local variablels:
    double  sx1, vx1;
    double    sy1, vy1;
    double    sz1, vz1;    
    
    // step to the next time in the simulation
    time+=tInc;

    // First calculate the direction cosines for the cannon orientation.
    // In a real game you would not want to put this calculation in this
    // function since it is a waste of CPU time to calculate these values
    // at each time step as they never change during the sim.  I only put them here in
    // this case so you can see all the calculation steps in a single function.
    b = L * cos((90-Alpha) *3.14/180);  // projection of barrel onto x-z plane
    Lx = b * cos(Gamma * 3.14/180);        // x-component of barrel length
    Ly = L * cos(Alpha * 3.14/180);        // y-component of barrel length
    Lz = b  * sin(Gamma * 3.14/180);    // z-component of barrel length

    cosX = Lx/L;
    cosY = Ly/L;
    cosZ = Lz/L;

    // These are the x and z coordinates of the very end of the cannon barrel
    // we'll use these as the initial x and z displacements
    xe = L * cos((90-Alpha) *3.14/180) * cos(Gamma * 3.14/180);
    ze = L * cos((90-Alpha) *3.14/180) * sin(Gamma * 3.14/180);
        
    // Now we can calculate the position vector at this time
    
    // Old position vector commented out:
    //s.i = Vm * cosX * time + xe;
    //s.j = (Yb + L * cos(Alpha*3.14/180)) + (Vm * cosY * time) - (0.5 * g * time * time);
    //s.k = Vm * cosZ * time + ze;

    // New position vector calc.:
    sx1 = xe;
    vx1 = Vm * cosX;
    
    sy1 = Yb + L * cos(Alpha * 3.14/180);
    vy1 = Vm * cosY;

    sz1 = ze;
    vz1 = Vm * cosZ;     

    s.i = ( (m/Cd) * exp(-(Cd * time)/m) * ((-Cw * Vw * cos(GammaW * 3.14/180))/Cd - vx1) -
          (Cw * Vw * cos(GammaW * 3.14/180) * time) / Cd ) -
          ( (m/Cd) * ((-Cw * Vw * cos(GammaW * 3.14/180))/Cd - vx1) ) + sx1;

    s.j = sy1 + ( -(vy1 + (m * g)/Cd) * (m/Cd) * exp(-(Cd*time)/m) - (m * g * time) / Cd ) +
          ( (m/Cd) * (vy1 + (m * g)/Cd) );

    s.k = ( (m/Cd) * exp(-(Cd * time)/m) * ((-Cw * Vw * sin(GammaW * 3.14/180))/Cd - vz1) -
          (Cw * Vw * sin(GammaW * 3.14/180) * time) / Cd ) -
          ( (m/Cd) * ((-Cw * Vw * sin(GammaW * 3.14/180))/Cd - vz1) ) + sz1;

Idealy if someone could "name" descriptively sx1, sv1, ... sz1, sv1, (even though I have a good idea that these are the initial translation and initial velocity.)

 

These are the initial variables:

	Vm		=	50;		// m/s
	Alpha	=	25;		// degrees
	Gamma	=	0;		// along x-axis
	L		=	12;		// m
	Yb		=	10;		// on x-z plane

	tInc	=	0.05;	// seconds
	g		=	9.8;	// m/(s*s)

	
	// Initialize the new variables:
	m		=	100;    // kgs
	Vw      =	10;		// m/s
	GammaW  =	90;		// degrees
	Cw      =	10;
	Cd      =	30;

I believe i'm after knowing Vm, Alpha and Gamma at the end (or probably at each step), and then how to use Alpha and Gamma to calculate the initial trajectory after the collision (say the collision plane is horizontal with its normal pointing straight up) (any CoR info would be useful too).

 

Many many thanks for any help ;)


#1rockpool

Posted 17 April 2013 - 10:09 AM

Hi,

I've got an example from a book working, however the formulas are so densely packed i'm having trouble really understanding it.

What i'm after is extracting the relevant information to extend the example and implementing a bounce.

The book is here, and the sample code can be found on the right, i'm using the Cannon 2 example:

http://shop.oreilly.com/product/9780596000066.do

 

Here's the DoSimulation function (it includes the preliminaries in each step for clarity)

    double    cosX;
    double    cosY;
    double    cosZ;
    double    xe, ze;
    double    b, Lx, Ly, Lz;
    double    tx1, tx2, ty1, ty2, tz1, tz2;

    // new local variablels:
    double  sx1, vx1;
    double    sy1, vy1;
    double    sz1, vz1;    
    
    // step to the next time in the simulation
    time+=tInc;

    // First calculate the direction cosines for the cannon orientation.
    // In a real game you would not want to put this calculation in this
    // function since it is a waste of CPU time to calculate these values
    // at each time step as they never change during the sim.  I only put them here in
    // this case so you can see all the calculation steps in a single function.
    b = L * cos((90-Alpha) *3.14/180);  // projection of barrel onto x-z plane
    Lx = b * cos(Gamma * 3.14/180);        // x-component of barrel length
    Ly = L * cos(Alpha * 3.14/180);        // y-component of barrel length
    Lz = b  * sin(Gamma * 3.14/180);    // z-component of barrel length

    cosX = Lx/L;
    cosY = Ly/L;
    cosZ = Lz/L;

    // These are the x and z coordinates of the very end of the cannon barrel
    // we'll use these as the initial x and z displacements
    xe = L * cos((90-Alpha) *3.14/180) * cos(Gamma * 3.14/180);
    ze = L * cos((90-Alpha) *3.14/180) * sin(Gamma * 3.14/180);
        
    // Now we can calculate the position vector at this time
    
    // Old position vector commented out:
    //s.i = Vm * cosX * time + xe;
    //s.j = (Yb + L * cos(Alpha*3.14/180)) + (Vm * cosY * time) - (0.5 * g * time * time);
    //s.k = Vm * cosZ * time + ze;

    // New position vector calc.:
    sx1 = xe;
    vx1 = Vm * cosX;
    
    sy1 = Yb + L * cos(Alpha * 3.14/180);
    vy1 = Vm * cosY;

    sz1 = ze;
    vz1 = Vm * cosZ;     

    s.i = ( (m/Cd) * exp(-(Cd * time)/m) * ((-Cw * Vw * cos(GammaW * 3.14/180))/Cd - vx1) -
          (Cw * Vw * cos(GammaW * 3.14/180) * time) / Cd ) -
          ( (m/Cd) * ((-Cw * Vw * cos(GammaW * 3.14/180))/Cd - vx1) ) + sx1;

    s.j = sy1 + ( -(vy1 + (m * g)/Cd) * (m/Cd) * exp(-(Cd*time)/m) - (m * g * time) / Cd ) +
          ( (m/Cd) * (vy1 + (m * g)/Cd) );

    s.k = ( (m/Cd) * exp(-(Cd * time)/m) * ((-Cw * Vw * sin(GammaW * 3.14/180))/Cd - vz1) -
          (Cw * Vw * sin(GammaW * 3.14/180) * time) / Cd ) -
          ( (m/Cd) * ((-Cw * Vw * sin(GammaW * 3.14/180))/Cd - vz1) ) + sz1;

Idealy if someone could "name" descriptively sx1, sv1, ... sz1, sv1, (even though I have a good idea that these are the initial translation and initial velocity.)

 

These are the initial variables:

	Vm		=	50;		// m/s
	Alpha	=	25;		// degrees
	Gamma	=	0;		// along x-axis
	L		=	12;		// m
	Yb		=	10;		// on x-z plane

	tInc	=	0.05;	// seconds
	g		=	9.8;	// m/(s*s)

	
	// Initialize the new variables:
	m		=	100;    // kgs
	Vw      =	10;		// m/s
	GammaW  =	90;		// degrees
	Cw      =	10;
	Cd      =	30;

I believe i'm after knowing Vm, Alpha and Gamma at the end (or probably at each step), and then how to use Alpha and Gamma to calculate the initial trajectory after the bollision (say the collision plane is horizontal with its normal pointing straight up) (any CoR info would be useful too).

 

Many many thanks for any help ;)


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