What length of rope do i need to climb the matterhorn?

swinging on a rope

• I know that the equation for using acceleration and velocity to determine position is: x(t) = a*a*t + v*t + x(0) I have a computer game where I have a guy that can move around and jump, etc. That equation is used to determine realistic movement. However, I want to add the capability to swing on a rope. Here are the values I know: x,y = where the character is holding the rope cX,cY = the center point that the rope is attached to r = the length of the rope g = the acceleration downward due to gravity t = the current time value I need an equation which I can use to determine the next point to display the character at. That means that given the values above I need to know: x(t),y(t) = the position of the character at that time vx(t),vy(t) = the velocity vector of the character at that time ax(t),ay(t) = the acceleration vector of the character at that time I need to know the velocities and accelerations because if the swing gets interrupted (i.e. they let go of the rope or swing into a wall) then I need to know what their acceleration and velocity vectors where at the time they were interrupted. I also need to be able to start the swing at any point in the arc. That means I could jump onto the rope while the rope is hanging straight down, or I could start swinging while the rope is pulled a ways to the side. I started with the equation for a half circle: x*x + y*y = r*r and messed around with the acceleration and velocity vectors and came up with an equation for the acceleration (who knows if it's right): Ax(t) = g*x(t) / sqrt( r*r - x(t)*x(t) ) but then I got stuck and wasn't able to figure out the rest.

• Answer:

  hi mxnmatch-ga, For uniformly accelerated motion your equation is x(t) = .5 *a*t*t + v*t + x(0) For swinging, motion is constrained to an arc, so that the only independent coordinate is the angle. In both cases, easiest way to display the motion is to use the energy conservation equation: 1/2 * m * v * v + m * g *h = constant m cancels out, of course g= 9.81 m / (s * s) is your a in free fall In both cases h is height above some level, e.g. above cy. For pendulum (polar coordinates) h or y = cy - r*sin(alpha(t)) x= cx + r*cos(alpha(t)) where alpha is angle from vertical The energy equation gives you v for every h you pick and polar coordinates above give you time t (or vice versa). from v and t you can get all other quantities, it is just trigonometry then. The diagram is shown at http://members.aol.com/calgea/l03eqatn.htm but you really need not to complicate your life with most of that. If you stick to the energy equation above and realize that the moment your guy lets go of the rope, the only acceleration is the gravity g. The speed vector vx, vy is given by the angle (theta in the drawing) and magnitude v (same as used above). The state of your guy is just [position, velocity]. The acceleration is gravity is known. So the state allows you to continue the motion to any next state, where you change the 'binding' [on the rope, in free flight]. Search Terms Pendulum, equation Please do rate my answer or ask if anything is not clear. hedgie