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Calculating the speed of a crane (using pythagoras) with different x & y speedsby Calico5
Tags: pythagoras theorem 
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#1
Dec2810, 09:19 AM

P: 2

Hi,
1. The problem statement, all variables and given/known data I'm designing a C++ program to simulate the operation of a steel processing warehouse. Part of my code requires me to calculate the time it takes for a crane to move certain distances (mainly diagonal ones) across a large shelving rack. 2. Relevant equations The crane moves at different speeds in different directions. Maximum x speed = 40m/min Maximum y speed = 12.5m/min I was thinking of using pythagoras theorem somehow to solves this (although its not required to be done this way) so: c^{2} = a^{2} +b^{2} 3. The attempt at a solution I was thinking of using: time = distance/speed to calculate the time for each movement in x direction and y direction. Then using pythagoras theorem on the times to get the diagonal movement time. So an example would be: XTime = 6/40 = 0.15min YTime = 4/12.5 = 0.32min ZTime = SQRT(0.15^{2}+0.32^{2}) = 0.353min Would that be correct? Im not sure if i can apply pythagoras theorem to time like that. Any Help is appreciated Thanks 


#2
Dec2810, 10:20 AM

P: 108

I would use Pythagorean theorem on the velocity and distance. Then use t = d/v. You would find how far the crane moves along the diagonal, and if the xy speeds are constant, how fast it is moving along the diagonal so you could calculate the time. 


#3
Dec2810, 11:46 AM

P: 2

Thank you for the reply Apphysicist :)
Ahh right, i see. Tried out your method and got the following: Z_{Distance} = SQRT(4^{2} + 6^{2}) = 7.211 Z_{Speed} = SQRT(40^{2} + 12.5^{2}) = 41.907 t=d/v = 7.211/41.907 = 0.17min That seems a bit quick doesnt it? given that the previous results were: XTime = 0.15min YTime = 0.32min i would have thought the pythagoras time would be closer to the slowest time. No? 


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