Tension in a wire Vs. force applied by a tensioned wire

Thread Starter

strantor

Joined Oct 3, 2010
6,798
if cable tension released, as in, say cable stretches, cylinders will move to account for this and pressure will resume its selected level.
Agreed.
the force applied will remain constant for a specific pressure and piston area. the only situation where angles would enter the picture would be in regard to side forces not in same plane and direction as piston movement
I don't follow. I'm using the term "Force applied" to describe that force which the advancing wire presses against the round object. Is that what you are describing here?
 

studiot

Joined Nov 9, 2007
4,998
OK I now understand, the device operates as a a giant fancy cheese or egg slice.

You are measuring the wire tension with a load cell and the pressure with some sort of gauge on the regulator.

How are you measuring the force on the pipe?
 

Kermit2

Joined Feb 5, 2010
4,162
Agreed.

I don't follow. I'm using the term "Force applied" to describe that force which the advancing wire presses against the round object. Is that what you are describing here?
I agree. If the object is round all side forces would resolve equally. wire contact with an object that was not symetrical in a right plane of the applied force would resolve with a resultant side force. i think you already deduced this.
 

Thread Starter

strantor

Joined Oct 3, 2010
6,798
Ok,
I've thought of (what I hope will be) a much better way to explain this:
Here is what I have now:
A wire stretched between 2 cliffs, with 5,000lbs of preexisting tension in it:
cliff1.png
Now I hang a mass of unknown weight from it, and the goal is to calculate the weight of it based on how much the cable deflects:
cliff2.png
I see that I now have 11,000lbs. of tension, and my cable has deflected 10 degrees. But how can I calculate the weight of the object? I think I would need to know things about the elasticity of the cable and rigidity of the cliffs to calculate it, and even then it probably wouldn't be very accurate. Am I right? am I missing something? Is there a way to do it?

And now here is what I WANT to do, by installing a pressure regulator on the cylinders:
To represent the pressure-regulated cylinders, I have inserted springs in the wire, and these springs can be considered as marvel-of-engineering-type (or magical-type) springs which have a constant tension; that is to say, that they will stretch to any length necessary to maintain 5,000Lbs. of tension.
Before hanging the weight:
cliff3.png
After hanging the weight:
cliff4.png
So I see here that tension, both before and after hanging the weight, is 5,000lbs. Preexisting tension is irrelevant. The weight of the object can be calculated like this:
Code:
Force_Applied_to_Object:= WEIGHT_OF_HANGING_OBJECT:= (Tension_Lbs)*(SIN(Left_Deflection_Angle)+SIN(Right_Deflection_Angle));
Force_Applied_to_Object:= WEIGHT_OF_HANGING_OBJECT:= (5000_Lbs)*(SIN(10_DEGREES)+SIN(10_DEGREES));
Force_Applied_to_Object:= WEIGHT_OF_HANGING_OBJECT:= (5000_Lbs)*(0.17365)+(0.17365));
Force_Applied_to_Object:= WEIGHT_OF_HANGING_OBJECT:= (5000_Lbs)*(0.3473); = 1736.5lbs
So, I've calculated that my object weighs 1736.5 Lbs. Does that look correct to you guys? Sound theory & math?
 

inwo

Joined Nov 7, 2013
2,419
Why is this? It seems to me that pressure in the cylinder should be directly proportional to tension in the wire, assuming a pressure regulator is used, and the cylinders are not bottomed out (somewhere between .
In the static simplified diagram shown, that's true.
Not all forces will be vertical (drawing) and measured by load cell.
I admit to not having a complete picture of things, obvious to others.

I should bow out. If you spend all your time correcting me, and teaching me, you won't get much else done. :(
So don't feel you must answer all my mistaken ideas.
I often post without enough thought............

It just seems to me that the forces can't be calculated from the simplified drawing.

The support structures of the driver and driven end seem all important to calculating the dynamic cable tension and cutting pressure.
 

studiot

Joined Nov 9, 2007
4,998
Your wire cutter is more akin to a bow and arrow than to a weight hung off a cliff.

Your analysis is too simple so how about answering my simple question in post#22?
 

Thread Starter

strantor

Joined Oct 3, 2010
6,798
OK I now understand, the device operates as a a giant fancy cheese or egg slice.

You are measuring the wire tension with a load cell and the pressure with some sort of gauge on the regulator.

How are you measuring the force on the pipe?
That's what I'm trying to calculate ;). There's no way to actually measure it while the device is in operation. I have measured it in the past using a setup like this:
3.png
Using temporary load cell #2 I have verified that the permanently installed load cell gives the correct tension for any angular offset of the trollies.
Using temporary load cell #1 I have verified that the formula: [Force_Applied_to_Object:= (Tension_Lbs)*(SIN(Left_Deflection_Angle)+SIN(Right_Deflection_Angle)) ] generates a force value that matches actual force applied, so long as I start with a slacked wire and drive in to take up the tension. - Keep in mind that I do not currently have the pressure regulator, and once I extend the cylinders, my dimensions (distances between sheave centers) become fixed values. So in order to test my Force_Applied formula, my test method was:
1. Test with cylinders fully retracted and wire completely slack - never apply any hydraulic pressure throughout the test.
2. Lengthen/shorten the wire with a come-along jack until it seemed like it would create the desired deflection angle once driven in (for example:10 degrees).
3. Drive the trollies in until slack is removed and 500lbs tension is registered on the permanently installed load cell and temporary load cell #2.
4. Measure the deflection angles (for example:10 degrees) and plug the numbers into my formula: [Force_Applied_to_Object:= (Tension_Lbs)*(SIN(Left_Deflection_Angle)+SIN(Right_Deflection_Angle)) ]
5. Record the reading of temporary load cell #1 and verify that the force measured is equal to the force calculated in step #4. (it was accurate, within 5-10% for every angle tested)
6. Retract the trollies until wire is completely slack
7. Let a little more slack out of the come-along jack until it looks like I will get another 10 degrees (for example:20 degrees)
8. Repeat procedure starting @ step #3; drive in until 500lbs tension, measure angle & calculate, verify accurate @ 20 degrees, 25 degrees, 30 degrees, 35 degrees, etc.

The above worked just fine when performed as described, but when I attempted the same test using hydraulic pressure in the cylinders (taut wire, fixed dimensions), the only deflection obtainable can be attributed to wire stretch and dimensional flex, and the maximum obtainable angle was about 9 or 10 degrees, and the result of my formula was WAY off.

I'm hoping that the regulator will allow my formula to be accurate in calculating force applied, as the cylinders will automatically give way, and deflection angle will automatically develop, with a known tension value, as the trollies are driven in and press against the object (or pull against temporary load cell #1).
 

studiot

Joined Nov 9, 2007
4,998
That's a big reply to a simple question.
I not imputing anything, just trying to gather data.

If you have not measured this force, how do you know your calculations are in error?
 

Thread Starter

strantor

Joined Oct 3, 2010
6,798
That's a big reply to a simple question.
I not imputing anything, just trying to gather data.

If you have not measured this force, how do you know your calculations are in error?
But I did measure it, using the setup & test procedure above.
...although, at the time, I was using the old (preexisting tension-actual tension) formula. perhaps it would have calculated the right tension had I been using the new formula that disregards preexisting tension.
 
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