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Old Jul 31st 2011, 03:02 AM   #1
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Force variation with increasing/decreasing distance and angle?

Practicle/Real Setup:
I have a glass screen of length 35" and width 46.5". I have mounted one force sensor at the left
top of glass screen and another sensor on the right top of the glass screen. Now, if I press my glass
screen from any point on the screen I shall get some force at the left and some force at the right sensor output accordingly.

Please refer to the attached figure.

I have hard and tought time to solve this problem. I pressed at point A with (x,y) = (3,32). At the left sensor I am
getting 7.015 Newtons. How much force I shall expect/get at right sensor?? Please refer the below info as well.

Point A with left sensor:
Angle = 45 degrees
Distance from sensor = 4.24"

Point B with right sensor:
Angle = 86.06 degrees
Distance from sensor = 43.6"

You can calculate angle and distance with your own calculation if I am wrong.

Thanks alot in advance,

Best Regards,
Naveed
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Force variation with increasing/decreasing distance and angle?-example.png  
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Old Aug 3rd 2011, 09:51 AM   #2
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I would have used the principle of moments... but I'm not sure if that would be applicable to 100%, especially if the glass sheet is not uniform. And I think that the mass of the glass sheet is important.

Anyway, you press at A with a force F, and as the weight of the glass is W, the total force at A becomes:

Sl = W/4 + (A)(32/35)([46.5-3]/46.5)

Sl is the share of force at the left sensor.

For the right sensor...

Sr = W/4 + (A)(32/35)(3/46.5)

Two equations with 3 variables, that's not possible to estimate the force registered by the right sensor unless the weight of the glass is known.

Also, a shortcoming in this 'experiment', when you press somewhere on the glass, does the glass remain dead horizontal? Because on the two other sides, the height of the corner above the ground might remain the same while the height of the corners at the sensors might decrease, meaning that the glass sheet is no more horizontal, and some of the weight of the glass sheet gets onto the two sensors, causing them to show a little more force than you expect.
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Old Aug 4th 2011, 06:28 AM   #3
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Thx alot dear,

I have uploaded the rough picture of screen for you to get some idea, sensors are in red color in picture , sensors are mounted between aluminum and glass screen. When you press glass it will press sensors.

Thx for the calculations, I have a question. We already have the value of left sensor which is 7.01N I got from sensor. So we only need to find the expected value at right sensor. I dont require exact/accurate force calculation. I want to find expected right sensor force(F2 or Fr) to do calibration for the right sensor by taking left sensor as a reference, As I am getting improper values as a result.

Thx alot for guidence,

Best Regards,
Naveed
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Force variation with increasing/decreasing distance and angle?-screen.png  
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Old Aug 4th 2011, 06:36 AM   #4
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Dear,

Just to update you, With my sensor's reading at when I pressed at point A. I got 7.01N at left sensor and 0.22N at right sensor. I know both values are wrong but left value looks bit ok!

So I want to calibrate right sensor. The application is not to measure the accurate pressure/force, the application is in touch sensing.

Your comments plz,

Regards,
Naveed
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Old Aug 4th 2011, 06:42 AM   #5
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Dear one query,

For the left sensor...
Sl = W/4 + (A)(32/35)([46.5-3]/46.5)

For the right sensor...
Sr = W/4 + (A)(32/35)(3/46.5)

If Sl is 7.01N , we can find "W" accordingly. Then A can also be found and finally Sr. But what is A?? Is it representing the point where we pressed(x,y)?

Naveed
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Old Aug 4th 2011, 06:44 AM   #6
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I m sorry, If A is known then W can be found. What is A?
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Old Aug 4th 2011, 07:53 AM   #7
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Well, if you have Sr and are sure about that, that should settle it, but your initial question was to find Sr (to then be able to calibrate it)

Oh and A is the force that you apply at A.

If you know Sl, you cannot find W, because substituting Sl, gives you:

7.01 = W/4 + (A)(32/35)([46.5-3]/46.5)

Simplifying...

7.01 = W/4 + (A)(32/35)([46.5-3]/46.5)

28.04 = W + 3.42A

And we are left with one equation and two unknowns, and another equation,

Sr = W/4 + (A)(32/35)(3/46.5)

with three unknowns.

(or simplified...)

4Sr = W + 0.236A

W is the one that I think is the easiest to find, that is the weight of the glass sheet, but since I don't know the circumstances of the experiment, I can't just say it like that

Ideally, I would like to know the absolute constants if I was in your place. The weight of the glass sheet won't change whatever happens, so I would be finding that first.

Otherwise, you can get A if you are using weights of specific and known weight.
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Old Aug 5th 2011, 10:11 PM   #8
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Dear ,

Thank you so much , the result looks ok!

but I have query and doubt

For left sensor
Sl = W/4 + (A)(32/35)([46.5-3]/46.5)

For right sensor...
Sr = W/4 + (A)(32/35)(3/46.5)

Can you explain plz that how did you drive this? I have question that left sensor's X-axis is 3" far from A so Sl shouldnt be
Sl = W/4 + (A)(32/35)(3/46.5)?
Sr = W/4 + (A)(32/35)([46.5-3]/46.5)?

Also explain that why did you add W/4 ?
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Old Aug 5th 2011, 10:13 PM   #9
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I mean the relation shall be

Sl = W/4 + (A)(32/35)(3/46.5)?
Sr = W/4 + (A)(32/35)([46.5-3]/46.5)?

Is it?
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Old Aug 6th 2011, 07:22 AM   #10
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I derived that from moments.

Take this, where X and Y are left sensor and Y is the right sensor.



W is the weight and A is the force applied.

From moments, we know that W is equally shared between X and Y, hence on X, you will have W/2 as force that W applies and on Y, you will have W/2 again. But knowing that there are four corners and that the weight acts through the centre of gravity at the geometric centre of the glass sheet, each corner will have first a force of W/4 due to the weight.

Now, when you consider the force A (in red), you have to work it out. The net moment is 0 (the sheet is not turning clockwise nor anticlockwise in the picture I uploaded. I will work with A only, and now, you can ingnore the W, since I already explained that to you and in all the calculations, you will have to include W/4 (which you shouldn't forget after the following steps)

So, clockwise moment about X gives you:
A*43.5 - Y*46.5 = 0

Clockwise moment due to A (A makes the sheet move clockwise) added with the moment due to Y (Y tends to 'counter' A, or make the sheet go counterclockwise, which is a reaction force. Remember Newton's 3rd law of motion, saying that every action has an equal and opposite reaction) equals the total moment, which is zero. Y is what you want. This gives: Y = 43.5A/46.5 = (43.5/46.5)A

Then you do the same about Y:
A*3 - X*46.5 = 0

Anticlockwise moment due to A added with clockwise moment due to X equals zero. X is what you want this time, and you get X = 3A/46.5 = (3/46.5)A

Note that this is in one dimension, you have to account for the other dimension because the (43.5/46.5)A represent all the force due to A only applied to the left side in your drawing, and (3/46.5)A represent all the force due to A only applied to the right side in your drawing.

You will further have to distribute that force between the upper and the lower corners, once with each part. Let's call (43.5/46.5)A = F1

Working through the same procedure, you should get the force at the top left corner as (32/35)F1. Substitute F1 and you get (32/35)(43.5/46.5)A.

And doing the same with the other part gives you (32/35)(3/46.5)A

If you want to get the other forces at the last two corners, you should get (3/35)(43.5/46.5)A for the lower left corner and (3/35)(3/46.5)A for the lower right corner.

Well, when you have done a lot of practice, you can grasp some shortcuts to all those calculations. For example, A is nearer to Y in my sketch. The only two values that I have are 43.5/46.5 and 3/46.5 from the lengths that I have. So, the force at Y is obviously the larger of the two, giving the force at Y (43.5/46.5)A. I repeat with the other.

And now, the TOTAL force at upper left corner is Sr = W/4 + (A)(32/35)(43.5/46.5) while at Sl, it's W/4 + (A)(32/35)(43.5/46.5)

I hope I didn't tell you too much at a time If you think you need explanation on something, let me know
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