Recent content by xovangam

and can i further simply that to: v = (g*d)/cos(a) * sqrt(2 * g * d * tan(a) + 2 * g * y) ?

cool. so A = (g*d)^2 B = cos(a)^2 C = -(2*g*d*sin(a))/(cos(a)) D = 2*g*y and thus v^2 = (g*d)^2/(cos(a)^2*(2*g*y + (2*g*d*sin(a))/cos(a)) ?

is this on the right track ? ((gd/v*cos(a)) - v*sin(a))^2 = v^2*sin(a)^2 + 2gy A = gd/v*cos(a) B = v*sin(a) (A - B)(A - B) = A^2 - AB - AB + B^2 = A^2 - 2AB + B^2 (gd/v*cos(a))^2 - 2*(gd/v*cos(a))*(v*sin(a)) + (v*sin(a))^2 = v^2*sin(a)^2 + 2*g*y => (gd)^2/(v*cos(a))^2 -...

given this formula for the range of a projectile when the initial height (y) is not zero: d = (v * cos(a) / g) * (v * sin(a) + sqrt((v * sin(a))2 + 2 * g * y) in a project I'm working on i need to compute 'v' given 'd', 'a', 'y' and 'g', so i would like to rewrite this equation in terms of 'v'...

i suppose i could also just do sin-1(Z) on the input for pitch to get theta now that i think about it..oh well. i agree it's more complex than it should be. in any case i think I'm all good now..

thanks @pmsrw3. to clarify i don't actually have theta when i calculate the launch direction, as i noted that comes from an external input as a number between 0 (no pitch) and 1 (straight up). so my givens are that and a target pos that is somewhere in front of me in the world. i will try...

i have a target in the world somewhere in front of me. i begin by calculating a displacement vector in the XY plane (Z is up) between the position i am going to fire from and the target: fireToTargetXY = (target_pos - fire_pos); fireToTargetXY.Z = 0.0; // i don't really need to do this because...

ah. so i have an initial launch direction that i start with (i.e. theta is known). well, more specifically, i have a unit vector that i start with. given that Z is up, i do this to calculate theta (pseudo code): up_vector = vector(0,0,1); Theta = acos(1.0 - (launch_direction dot up_vector))...

@pmsrw3: the initial velocity is being calculated as: v = sqrt((d * g) / sin (2 * theta)) as i simply solved for v in the initial equation. is that not correct ? @robphy: sorry, i failed to mention that Z is "up" in my 3D world, so Z is the vertical component, X and Y are horizontal...

actually, using smaller timestep increments doesn't seem to bring me closer to the target.. is d = (v2 / g) * sin(2 * theta) not what i want to be using here ?

hi there, i'm programming a simulation of a projectile that has a known target off in the distance somewhere in my 3D world, i also know the angle at which i will be firing the projectile. i am firing the projectile on flat ground (so the start and end y-coordinates are the same). i'm...

very cool. thank you sir you've been more than helpful!

ok, so to summarize the results, the solution proposed by micromass worked (thanks!), but i had to negate the (xi',yi') result for any point (xi, yi) inside the larger rectangle that i transformed. i *think* this is because the coordinate system for the larger rectangle actually does have the...

that almost works :smile: if Q in the larger rectangle is (-x,-y) however, it's coming out as (-x',-y'), which is outside the boundary of the smaller rectangle. though the values seem correct, they are just negated to the opposite (-x,-y) quadrant it looks like. maybe i set it up wrong, i...

isn't this a change of basis though? disregarding scale, i guess I'm not clear on how i could rotate the larger rectangle within the standard R2 space and end up with the coordinate system of the smaller rectangle (where positive Y is flipped to point down with positive X still pointing to the...