Jump to content

  • Log In with Google      Sign In   
  • Create Account

FREE SOFTWARE GIVEAWAY

We have 4 x Pro Licences (valued at $59 each) for 2d modular animation software Spriter to give away in this Thursday's GDNet Direct email newsletter.


Read more in this forum topic or make sure you're signed up (from the right-hand sidebar on the homepage) and read Thursday's newsletter to get in the running!


#ActualHappyCoder

Posted 17 April 2013 - 11:24 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.

#3HappyCoder

Posted 17 April 2013 - 11:23 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.

#2HappyCoder

Posted 17 April 2013 - 11:22 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.
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.

#1HappyCoder

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.
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 * unitX;

PARTNERS