Calculating the speed of a crane (using pythagoras) with different x & y speeds

Calico5
Messages
2
Reaction score
0
Hi,

Homework Statement



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.

Homework 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:
c2 = a2 +b2

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:

14ybc5s.jpg


X-Time = 6/40 = 0.15min
Y-Time = 4/12.5 = 0.32min

Z-Time = SQRT(0.152+0.322) = 0.353min

Would that be correct? I am not sure if i can apply pythagoras theorem to time like that. Any Help is appreciated

Thanks
 
Physics news on Phys.org
Calico5 said:
Hi,

Homework Statement



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.

Homework 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:
c2 = a2 +b2

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:

14ybc5s.jpg


X-Time = 6/40 = 0.15min
Y-Time = 4/12.5 = 0.32min

Z-Time = SQRT(0.152+0.322) = 0.353min

Would that be correct? I am not sure if i can apply pythagoras theorem to time like that. Any Help is appreciated

Thanks


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 x-y speeds are constant, how fast it is moving along the diagonal so you could calculate the time.
 
Thank you for the reply Apphysicist :)

Ahh right, i see. Tried out your method and got the following:

ZDistance = SQRT(42 + 62) = 7.211
ZSpeed = SQRT(402 + 12.52) = 41.907

t=d/v = 7.211/41.907 = 0.17min

That seems a bit quick doesn't it? given that the previous results were:
X-Time = 0.15min
Y-Time = 0.32min

i would have thought the pythagoras time would be closer to the slowest time. No?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top