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gimpy

Simple Question.

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Ok i just want to comfirm that i did this correctly. I am making a simulator for a missile launch. Anyways the Launch pad is located at 0m and the target is located 250m away and it is 170m high and flat and infinitly long. Now i want the missile to just display where it hits on the target (170m high, if it can fit on the screen or course). So this is what i did. I used for the x and y positions x = t*V*cos(alpha) y = t*V*sin(alpha) - (g*t^2)/2 t = ? V = 200 feet per second g = 9.81 alpha = 45 degrees * PI/180 (conversion to radians) Now i know what y is (170m) so to find x (impact point) i must solve the equation y = t*V*sin(alpha) - (g*t^2)/2 for t then substitute t into x = t*V*cos(alpha). (gt^2)/2 - t*V*sin(alpha) + y = 0 so i solve this using the quadratic equation (-b +- sqrt(b^2 - 4*a*c))/2*a where a = g/2 b = -(V*sin(alpha)) c = y to obtain the roots , which are the times in seconds where the missile is at h=170m (going up and coming down). So obviously the the greater root (t) is the impact point of t (where the projectile lands on the target). Oh and to solve the quadratic equation i divided a, b and c by a, just to make a = 1. ( i get a different answer if i don't ) Does this sound right? Im pretty sure it is, i just want to make sure. Oh an i coded this using Java, no openGL or anything, just simple GUI that will display the impacy point on the target (yeah i know it's kinda lame but thats what i am required to do). Thanks Stupid Everything!! [edited by - gimpy on March 18, 2004 1:35:07 PM]

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