• Content count

  • Joined

  • Last visited

Community Reputation

0 Neutral

About frumpydolphin

  • Rank

Profile Information

  • Favorite Area of Science
  1. Well, I can't think of a good way to explain so sorry for time waste. Thanks for trying to help though.
  2. Geez I suck at this, a, b, and c act as the origin of the vectors, the direction is the direction of the vector from a, b, or c towards q
  3. Same time
  4. Ya, so there is a point(q) starting with coordinates x​q and yq​. Three other points a, b, and c have there x and y coordinates labelled in the same manner​. A, b, and c exert a "push" on q each second suppose, and move it to the endpoint of a vector 2 times the magnitude of vector a,b, or c to q. The direction remains the same and the origin of the vector is still a,b or c. The average of the endpoints created by these vectors, 1 for a, 1 for b, and1 for c, gives q's position after 1 second. After n times though q's position would be the average of all these averages. I derived the sum to n from 1 of the small averages divided by the number of seconds, so n, to get the final x and y values of q. (really need to figure out how to use equations) Hope that helps but that is only the first part.
  5. Background - System of points a, b, c and q - a, b, and c exert a push on q moving it to the end of a vector that is 2 times the vector from a, b, or c to q. - a, b, and c do this at the same time, after each cycle the x and y values of q are the average of the end points of the vectors. - This happens n times First derivation The final x value is the sum of the averages of all the vectors endpoints divided by the number of iterations so n. Same goes for y I further thought why keep it at intervals when it can be a constant force. From this I derived The integral from 1 to n of the average of the averages of the vector endpoints in respect to the x value of q. Same for y Hope I wasn't too vague with it but am I at least close to right? Also how do I insert equations? This wasn't homework by the way don't rage at me.