Martin,

*>Cantenary Curve Effect*

*><http://www.spaceagecontrol.com/calccabl.htm> *

Looks like what I am looking for, Thanks.

Spread-sheet time.

Curbie

Mike,

*> I can't help you with the math, but you might ask on sci.physics.*

Thanks, I give it a try.

Curbie

Curbie wrote:

*> *

*> Any idea on terms for keywords I should use to search for the math?*

*> *

*> Thanks,*

*> *

*> Curbie*

*> *

Try "Triangle of forces"

Tim

Curbie wrote:

*> Jim,*

*> *

*>> Here's a commercial one. Click on the Tension Meter (long URL):*

*>> http://www.dillon-force.com/ *

*> Thanks, they have a large image of their device showing mechanical*

*> details here:*

*>*

*
http://www.dillon-force.com/components/com_virtuemart/shop_image/product/abdb983a1fa6ad01da4c05c327dcb7d5.jpg *
*> *

*> The device seems to use the same principles (math) for cables that the*

*> device I posted uses for wire, and I'm still interested in the math*

*> behind it.*

*> *

*> Any idea on terms for keywords I should use to search for the math?*

*> *

*> Thanks,*

*> *

*> Curbie*

*> *

Well the 'math' goes along these lines (pun intended:-).

Looking at the straight line where the wire is originally and the wire

position after it's deflected, and draw a line perpendicular from the

mid-point over to the original line of the wire. We have a right

triangle with one leg's length 1/2 inch and the other leg's length 20

inches and the hypotenuse is the wire. With such a long narrow

triangle, we can approximate the length of the hypotenuse as also being

20 inches. And the sin of that narrow angle is the opposite over

hypotenuse so we have (0.5)/20 = 1/40.

The tension in the wire is pulling at a very shallow angle, trying to

straighten out. So the sideways force created by the tension in the

wire is X*sin(angle) = X/40.

Looking at the whole thing you can see that it is symmetrical and there

is a second triangle formed on the other side of the point where the

scale attaches. The tension in *that* segment of wire is also 'X' and

it also is creating a side force of X/40.

So these two sections of wire are creating a side force of X/40 + X/40

or X/20. And that side force is exactly countered by the pull on the scale.

So working it the other way around, the pull on the scale times 20

equals the tension in the wire.

daestrom

*> ...So the sideways force created by the tension in the*

*> wire is X*sin(angle) = X/40....*

*> daestrom*

Nice, simple explanation. I couldn't figure out how to make vector

force resolution intuitively obvious.

jsw

>Cantenary Curve Effect><http://www.spaceagecontrol.com/calccabl.htm>