48 replies to this topic

[Matt]

But a cable suspended between two points does not follow a parabola, as the function y=10^(-3)(x^2) does. A catenary curve is not a parabola. I don't know how to do it if the two end points are at different heights. The curve of a cable suspended between two points is y = a*cosh(x/a) where cosh is hyperbolic cosine. The length of the curve is given by a*sinh(x/a). The parabola is an approximation of the catenary curve, and the equation CAski gave for the length is also an approximation.

[CAski]

Yes, the parabolic equation was just a random approximation. However, L =∫(√(1+(dy/dx)^2))dx over the horizontal distance is most certainly not an approximation and works no matter what your equation f(x) is. Your equations appear to be correct, and I believe that the Arc Curve equation you gave is just a simplification of L =∫(√(1+(dy/dx)^2))dx where f(x)=acosh(x/a).

[CAski]

Oh! This has been most exciting! Cosh and Sinh are actually a delight to work with the Length equation I have given. Using the Length equation, I came with a proof that the length for an equation f(x)=acosh(x/a) can be simplified to L=asinh(x/a). I really haven't felt this happy with mathematics in a very long time.

[CAski]

Here it is! The tricky part was simplifying 1+1/4(e^(x/a)-e(-x/a))^2 to make the equation work!

y=acosh(x/a)
y=1/2a(e^(x/a)+e(-x/a))
y=1/2ae^(x/a)+1/2ae(-x/a)

dy/dx=1/2a(1/a)e^(x/a)+1/2a(-1/a)e^(-x/a)
dy/dx=1/2e^(x/a)-1/2e^(-x/a)
dy/dx=1/2(e^(x/a)-e^(-x/a))

(dy/dx)^2=1/4(e^(x/a)-e^(-x/a))^2

L=∫(1+(dy/dx)^2)^(1/2)dx
L=∫(1+1/4(e^(x/a)-e^(-x/a))^2)^(1/2)dx
L=∫(1+1/4(e^(2x/a)-2+e^(-2x/a)))^(1/2)dx
L=∫(1+1/4e^(2x/a)-1/2+1/4e^(-2x/a))^(1/2)dx
L=∫(1/4e^(2x/a)+1/2+1/4e^(-2x/a))^(1/2)dx
L=∫(1/4(e^(2x/a)+2+e^(-2x/a)))^(1/2)dx
L=∫(1/4(e^(x/a)+e^(-x/a))^2)^(1/2)dx
L=∫1/2(e^(x/a)+e^(-x/a))dx
L=∫(1/2e^(x/a)+1/2e^(-x/a))dx
L=1/2ae^(x/a)-1/2ae^(-x/a)+c

L=1/2a(e^(x/a)-e^(-x/a))+c

L=asinh(x/a)

Wow. Math appearantly was not meant to be posted in forums. That is pretty dense. However, the point is, Matt, that it should now be pretty clear that the equation I presented is not only not an approximation, but that it is the device through which the equation you presented was derived in the first place.

[Matt]

The precise definition of the length of the curve isn't of much use if you don't have the correct function defining the curve. Even Galileo believed adamantly that a sagging cable forms a parabola.

Thank you for the proof. I know it must have been labour intensive. I'll spend a while trying to figure that one out.

Also remember that the catenary curve equation is an approximation, albiet, an extemely accurate one, of a more complex function. The other assumptions we are making in the approximation are that the cable has absolutely no rigidity, and is complety free to pivot at the ends.

[Dawson]

Wow, nice work by everyone one this thread!!

When a lift is stationary (ie static) the tension will be the same everywhere on the cable. This is because there are no losses due to friction etc, and by definition for it to be static there can be no net forces. This is regardless of whether the uphill side is loaded or not.

When the lift is moving, the cable tension is highest in the reach of cable between the bullwheel and the first sheave battery. Losses due to friction etc mean that the cable tension progressively looses tension across every battery back to the drive again. The difference in the tension between the two lengths of cable on either side of the bullwheel is equal to the force imparted on the cable by the drive. I'm the wrong kind of engineer to know if this is 100% correct, so I'm happy to hear any thoughts. I think the difference in tension on the uphill and downhilll sides of the return is resolved by the rotational momentum in the return bullwheel.

If you take all the brakes off the lift then it will move if there is enough load to overcome the friction losses. Obviously a sand bag is not enough load. The load also needs to be on a cable span with some grade - ie. the span into the load and unload are horizontal so aren't going to cause cable movement.

[liftmech]

I think you're incorrect on one point. The tension will always be higher at the top terminal, regardless of whether the lift is moving. At the bottom there is only the length of rope between the first sheaves and the bullwheel, and any carriers that might be in that span. At the top the bullwheel is being acted upon by the full weight of the line, including carriers.

[nathanvg]

Dawson, on Jun 21 2005, 10:43 PM, said:

Wow, nice work by everyone one this thread!!

When a lift is stationary (ie static) the tension will be the same everywhere on the cable.  This is because there are no losses due to friction etc, and by definition for it to be static there can be no net forces.  This is regardless of whether the uphill side is loaded or not.

When the lift is moving, the cable tension is highest in the reach of cable between the bullwheel and the first sheave battery.  Losses due to friction etc mean that the cable tension progressively looses tension across every battery back to the drive again.  The difference in the tension between the two lengths of cable on either side of the bullwheel is equal to the force imparted on the cable by the drive.  I'm the wrong kind of engineer to know if this is 100% correct, so I'm happy to hear any thoughts.  I think the difference in tension on the uphill and downhilll sides of the return is resolved by the rotational momentum in the return bullwheel.

If you take all the brakes off the lift then it will move if there is enough load to overcome the friction losses.  Obviously a sand bag is not enough load.  The load also needs to be on a cable span with some grade - ie. the span into the load and unload are horizontal so aren't going to cause cable movement.
<{POST_SNAPBACK}>

Dawson, I think you have oversimplified the problem and there are a few errors in your logic as a result:

>> When a lift is stationary (ie static) the tension will be the same everywhere on the cable

This would only be true if the lift was level (no vertical rise), all wheels(including sheaves) in the system did not have any static friction, distribution of carrier weight was even, distances between lift poles were even and possibly other factors too. You are correct that the net force on any one point will be zero but even though the lift is stopped there are forces other than tension acting on the system. These forces include static friction and the force of gravity. These factors will allow varying tension at various sections of the cable.

>> When the lift is moving, the cable tension is highest in the reach of cable between the bullwheel and the first sheave battery.

This might be true but once again the force of gravity and friction could make this incorrect.

EXAMPLE: Imagine a tram with two very heavy trams that is very steep. The tension at the top would always be very high to overcome the weight of the two trams. If the drive wheel was at the bottom and friction was low, the drive could have much lower tension.

[Dawson]

Ah, of course. My bad. Just like a rope hanging with weights evenly spaced along in, the tension increases with each weight. No wonder the chairlift I built out of leggo didn't work