Homework Help: Parabola problem - some data given.

1. Aug 11, 2012

yoghu

1. The problem statement, all variables and given/known data

Essentially there is a power line which must be strung between posts on two mountains which are 1669.602m apart.

The first mountain is 5m higher than the second.

To overcome tension the wire is strung with a sag, 1m of sag for every 75m, or part thereof. This sag is relative to the lower pylon.

Determine an equation for the profile of the wire between the two mountains, if the arc is to be in a PARABOLIC shape.

2. Relevant equations

I've worked out that 1669.602/75 = (22. something) which means there must be 23m of sag, as it states part thereof. Relative to the lower pylon, I understand this to mean the turning point will be 23m below the smaller mountain.

Due to the difference in heights, this is obviously not a symmetrical parabola.

3. The attempt at a solution
I've tried letting the turning point be the origin, and having 3 points
(X2 - 1669.602 , 28)
(1669.602-X1, 23)
(0,0)
where x2 is the distance from the turning point to the second mountain, and X1 is the distance between the first mountain to the turning point.

I've also tried letting the first mountain be the origin, with 3 points
(0,0)
(???,-28) - minimum point
(1669.602,-5)

But I'm very stuck on what to do with these possible points. Please help.

Last edited: Aug 11, 2012
2. Aug 11, 2012

phinds

but the arc will NOT be a parabola ! It will be a catenary.

3. Aug 11, 2012

yoghu

I know, but the question specifically asks for a parabolic equation.

It also asks to discuss limitations of this shape, so that's where I'd bring in your point.

4. Aug 11, 2012

phinds

But how can you produce an equation for a parabola to describe a curve that isn't a parabola? What would be the POINT of such an exercise?

5. Aug 11, 2012

yoghu

I have no idea, and I don't know why.

Do you have any idea on how to do it??

6. Aug 11, 2012

Staff: Mentor

It's a mathematical exercise in curve fitting. The mountain/powerline theme is just window dressing that provides a minds-eye picture of the setup.

7. Aug 11, 2012

phinds

So is one then supposed to first find the correct curve (the catenary) and then use that do find the closest possible parabola curve? Seems like an amazing waste of time. What am I missing?

8. Aug 11, 2012

Staff: Mentor

There is enough information to fit a parabola; consider that you effectively have three given points through which the curve passes.

9. Aug 11, 2012

phinds

DOH ! NOW I see what you meant in post #6. I'm just overcomplicating things. Thanks for the clarification.