Posted by Curbie on July 8, 2009, 2:51 am
wrote:
>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
daestrom,
Thanks for the easy to understand explaination; it's going to save me
a lot of time!
Mike & Tim had me headed down the right track, thank you two also.
Curbie
Posted by Richardson on July 6, 2009, 3:33 am
> This guy outlines a simple tension meter you can build it yourself for
> about $.00
>
http://www1.foragebeef.ca/ $foragebeef/frgebeef.nsf/all/frg35/$FILE/fencetension.pdf
> It's for wire and not cable in which I'm interested in but the idea is
> pretty simple:
> Two wire guides 40" apart, deflect the wire with a spring scale ½
> inch, multiply the scale reading by 20 to determine the amount of
> tension on the wire.
> I know this has no direct value with cables with 1000's of pounds of
> tension (like guys), but I was wondering if anyone knows or can point
> me to the math behind the idea?
> Thanks,
> Curbie
Curbie,
Let me teach you something, Motor and Generator use Permanent magnet are
the same. That's what they use blocking diode to control the current flow.
Don't be picky, learn the core of science and stop fooling around with the
two dead animals in this group or you will be infected like them.
Regards.
Posted by daestrom on July 7, 2009, 10:26 pm
Richardson wrote:
>> This guy outlines a simple tension meter you can build it yourself for
>> about $.00
>>
http://www1.foragebeef.ca/ $foragebeef/frgebeef.nsf/all/frg35/$FILE/fencetension.pdf
>>
>> It's for wire and not cable in which I'm interested in but the idea is
>> pretty simple:
>>
>> Two wire guides 40" apart, deflect the wire with a spring scale ½
>> inch, multiply the scale reading by 20 to determine the amount of
>> tension on the wire.
>>
>> I know this has no direct value with cables with 1000's of pounds of
>> tension (like guys), but I was wondering if anyone knows or can point
>> me to the math behind the idea?
>>
>> Thanks,
>>
>> Curbie
>>
>
>
>
> Curbie,
>
> Let me teach you something, Motor and Generator use Permanent magnet are
> the same. That's what they use blocking diode to control the current flow.
> Don't be picky, learn the core of science and stop fooling around with the
> two dead animals in this group or you will be infected like them.
>
Ah, you've proven that you can't read the original poster's message. (or
you can't figure out how to use your news reader)
daestrom
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>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