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Physics for Game Developers book example help


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#1 rockpool   Members   -  Reputation: 103

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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 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 ;)


Edited by rockpool, 17 April 2013 - 10:10 AM.


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#2 HappyCoder   Members   -  Reputation: 2887

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Posted 17 April 2013 - 11:21 AM

    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
This part takes two angles and converts then into a vector.
Gamma appears to be your yaw, Alpha appears to be your pitch.
When Alpha is 0 the cannon is pointing straight up. When Alpha is 90 the cannot is horizontal.
When Gamma is 0 the cannon points in the positive X direction and as it increases it starts to point in the positive Z direction.

I don't know why they use cos(90-Alpha) That is the same as putting sin(Alpha)
    cosX = Lx/L;
    cosY = Ly/L;
    cosZ = Lz/L;
This part calculates unit vector of L. It goes in the same direction as L but is only of length 1. cosX, cosY, cosZ aren't really good variable names where.
    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);
The same as Lx and Lz. This is just a recalculation. This could really just be xe = Lx and ze = Lz.
    sx1 = xe;
    vx1 = Vm * cosX;
    
    sy1 = Yb + L * cos(Alpha * 3.14/180);
    vy1 = Vm * cosY;
 
    sz1 = ze;
    vz1 = Vm * cosZ;
Sets s_1 to be the position at the end of the barrel. And sets v_1 to be the velocity moving in the direction of the barrel. Remember cos_ actually represents the unit vector in the direction of the barrel. By multiplying it by Vm you get a vector in the same direction only with the length of Vm.
   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;
This calculates the current position of a projection based on the starting position, starting velocity, and time.



Honestly this code is pretty bad. It is hard to read, i recalculates values all over and doesn't have good variable names. If you are just after the velocity then this might be more useful
#define DEG_2_RAD(deg) (deg * 3.14/180)


double unitX = sin(DEG_2_RAD(Alpha)) * cos(DEG_2_RAD(Gamma));
double unitY = cos(DEG_2_RAD(Alpha));
double unitZ = sin(DEG_2_RAD(Alpha)) * sin(DEG_2_RAD(Gamma));

double velocityX = StartSpeed * unitX;
double velocityY = StartSpeed * unitY;
double velocityZ = StartSpeed * unitZ;

double startX = BarrelLength * unitX + cannonX;
double startY = BarrelLength * unitY + cannonY;
double startZ = BarrelLength * unitZ + cannonZ;
velocity_ is the start velocity. start_ is starting position of the projectile. I would also recommend either finding or making a vector class. A vector class would let you calculate all three components at once instead of calculating each on in a separate line. So calculating the speed would be Vector3 velocity = StartSpeed * unit; where unit is also of type Vector3.

Edited by HappyCoder, 17 April 2013 - 11:24 AM.


#3 rockpool   Members   -  Reputation: 103

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Posted 17 April 2013 - 11:40 AM

So what is Gamma, Alpha and the velocity at any given time? I believe you've only rewritten the first part...

 

TBH, i've rewritten it in Lua so no vector3's...



#4 rockpool   Members   -  Reputation: 103

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Posted 17 April 2013 - 12:46 PM

1. If I subtract the last position from the current position position, would that vector be its current velocity?

 

2. If I used some trig to calculate the angle between the last and current position, would that satisfy the Alpha and Gamma?

 

 

 

1.1 Negating deformation, if the ball hits a surface with a CoR of say 0.9 and the ball has a CoR of say 0.92, will I use an average of the two, e.g. 0.91 * the velocity from 1?

 

2.1 Negating spin, if I have the approach angle relative to the normal of the surface, will the exiting angle be a reflection of this angle of incidence? (or was that just a drunken night playing pool and seeing double?)... and how to do using two vectors?


Edited by rockpool, 17 April 2013 - 12:49 PM.


#5 DT....   Members   -  Reputation: 487

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Posted 17 April 2013 - 03:06 PM

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

 

 

http://www.gamedev.net/topic/204055-calculating-reflection-vector-i-have-no-math-book-here/



#6 rockpool   Members   -  Reputation: 103

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Posted 17 April 2013 - 03:10 PM

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

 

 

http://www.gamedev.net/topic/204055-calculating-reflection-vector-i-have-no-math-book-here/

Cheers, that's a lot simpler than reading this (at the bottom):

http://www.gamasutra.com/view/feature/131424/pool_hall_lessons_fast_accurate_.php?page=3






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