How to find coordinates of 4 points (shown in picture: thermoelectric cooler device)

[nayminsit]

Please help me!

I have a problem as shown in figure. Actually it is a thermoelectric cooler device. But here i need to find only coordinates points of last block coner 4 points mathematically (in figure i show with RED sport). My Supervisor gave me experimental result from factory test (sizes of device and after testing result) in figure i shown. how can i calculate? please explain me.

 

[Dr.Peterson]

View attachment 10275


Please help me!

I have a problem as shown in figure. Actually it is a thermoelectric cooler device. But here i need to find only coordinates points of last block coner 4 points mathematically (in figure i show with RED sport). My Supervisor gave me experimental result from factory test (sizes of device and after testing result) in figure i shown. how can i calculate? please explain me.

The answer would depend on the shape of the curve. Is it known to be a semi-ellipse, as it appears to be, or is that just the way you happened to draw it for convenience, and it's really something else, such as a circular arc or a catenary or some unnamed curve?

 

[nayminsit]

The answer would depend on the shape of the curve. Is it known to be a semi-ellipse, as it appears to be, or is that just the way you happened to draw it for convenience, and it's really something else, such as a circular arc or a catenary or some unnamed curve?

Thank you sir!

The shape after test is little curve 13x10^-6 meter and length elongated 5x10^-6 meter. So my supervisor advise me to imagine like a circle and first need to find radius and then to find position of last edge’s 4 points. Now it is hard for me to find even radius. I think it will need to use (radius of curvature) but he said he needs only the positions and nothing is important next. Radius of curvature is difficult to understand for me. If you have any idea to solve please help me.

 

[Dr.Peterson]

Thank you sir!

The shape after test is little curve 13x10^-6 meter and length elongated 5x10^-6 meter. So my supervisor advise me to imagine like a circle and first need to find radius and then to find position of last edge’s 4 points. Now it is hard for me to find even radius. I think it will need to use (radius of curvature) but he said he needs only the positions and nothing is important next. Radius of curvature is difficult to understand for me. If you have any idea to solve please help me.

Okay, so the picture is not much like the reality; it's really more like this:


But as you may have noticed, your picture is too small to read most of the dimensions; and what I can read makes no sense -- the width of the whole thing after bending, 5*10^-6 m, appears to be far smaller than the width of just part of it before, 1.4 mm. Also, the inner and outer arc lengths have to be different, so at least one of them has to have changed from the original length, and we would need to know how -- does one stay the same, or is their average unchanged, or what?

Once we have the dimensions, the width and height together with the arc length(s) can give us the radius and the angle, using formulas here:
http://mathworld.wolfram.com/CircularSegment.html
http://mathforum.org/dr.math/faq/formulas/faq.circle.html#segment

There will be one more issue: when you refer to coordinates, where do you want to measure them from? What axes do you want to use, or did you really mean some other kind of dimensions?

 

[nayminsit]

Okay, so the picture is not much like the reality; it's really more like this:

View attachment 10279
But as you may have noticed, your picture is too small to read most of the dimensions; and what I can read makes no sense -- the width of the whole thing after bending, 5*10^-6 m, appears to be far smaller than the width of just part of it before, 1.4 mm. Also, the inner and outer arc lengths have to be different, so at least one of them has to have changed from the original length, and we would need to know how -- does one stay the same, or is their average unchanged, or what?

Once we have the dimensions, the width and height together with the arc length(s) can give us the radius and the angle, using formulas here:
http://mathworld.wolfram.com/CircularSegment.html
http://mathforum.org/dr.math/faq/formulas/faq.circle.html#segment

There will be one more issue: when you refer to coordinates, where do you want to measure them from? What axes do you want to use, or did you really mean some other kind of dimensions?

Thank you Sir!
Yes, your picture is exactly like he wants!!!
I will show you his hand drawing pictures and the Normal sate of my device (means dimensions) in large picture.
No (1) pic is original dimensions (Normal State)
N0 (2) pic is horizontal view (compared before and after)
N0 (3) pic is vertical view (compared before and after)
N0 (4) pic is he adviced (like your pic)

(1)

(2)

(3)

(4)

I think 13x10^-6 m curved and 5x10^-6 m elongated are only the different between top surface line.
Yes, deformation is so small. measured with some special devices in factory. Our goal is when we know and prove this deformation process and then we will create better with Physical Ways.

 

[nayminsit]

Okay, so the picture is not much like the reality; it's really more like this:

View attachment 10279
But as you may have noticed, your picture is too small to read most of the dimensions; and what I can read makes no sense -- the width of the whole thing after bending, 5*10^-6 m, appears to be far smaller than the width of just part of it before, 1.4 mm. Also, the inner and outer arc lengths have to be different, so at least one of them has to have changed from the original length, and we would need to know how -- does one stay the same, or is their average unchanged, or what?

Once we have the dimensions, the width and height together with the arc length(s) can give us the radius and the angle, using formulas here:
http://mathworld.wolfram.com/CircularSegment.html
http://mathforum.org/dr.math/faq/formulas/faq.circle.html#segment

There will be one more issue: when you refer to coordinates, where do you want to measure them from? What axes do you want to use, or did you really mean some other kind of dimensions?

I found some beam bending problem. Please check!

 

[Dr.Peterson]

I found some beam bending problem. Please check!

View attachment 10284

That looks related, but it doesn't answer any of my questions.

 

[Dr.Peterson]

Thank you Sir!
Yes, your picture is exactly like he wants!!!
I will show you his hand drawing pictures and the Normal sate of my device (means dimensions) in large picture.
No (1) pic is original dimensions (Normal State)
N0 (2) pic is horizontal view (compared before and after)
N0 (3) pic is vertical view (compared before and after)
N0 (4) pic is he adviced (like your pic)

(1)
View attachment 10280
(2)
View attachment 10281
(3)
View attachment 10282
(4)
View attachment 10283
I think 13x10^-6 m curved and 5x10^-6 m elongated are only the different between top surface line.
Yes, deformation is so small. measured with some special devices in factory. Our goal is when we know and prove this deformation process and then we will create better with Physical Ways.

I saw your other reply before I saw this one. This does answer some of my questions.

I figured out that your ∆l, which was shown as the actual width on the picture, is supposed to be only the change in width. So apparently the chord is 40 mm + 5×10^-6 m (though I'd expect it to be shortened rather than lengthened!), while the height of the arc (sagitta) is 13×10^-6 m. I still don't know whether to think of the arc length as identical to the original length, but we don't need to know that. In fact, I'll ignore the ∆l, as it is insignificant in terms of significant digits.

Using the formulas from http://mathforum.org/dr.math/faq/faq.circle.segment.html#8,

Case 8: You know c and h. Then

r = (c²+4h²)/(8h),
theta = 2 arcsin(c/[2r]),
s = r theta,
d = r - h,
K = r²[theta-sin(theta)]/2.​

and taking c = 40×10^-3 m and h = 13×10^-6 m, we can find the radius,

r = (c^2 + 4h^2)/(8h) = (1600×10^-6 + 4×169×10^-12)/(8×13×10^-6) = 15.38 m
​

The angle subtended by the arc is

theta = 2 arcsin(c/[2r]) = 2 arcsin(40×10^-3/[2×15.38]) = 0.149°
​

Check me on the calculations!

Now, to find coordinates, you can use a little trig. The angle along the arc will be proportional to the distance along the length in the unbent form (presumably), so you can work out the angle along the arc, and use sine and cosine to find the coordinates. I don't think anything you sent shows how you want those coordinates measured, so I'm not ready to do that for you.

 

[nayminsit]

I saw your other reply before I saw this one. This does answer some of my questions.

I figured out that your ∆l, which was shown as the actual width on the picture, is supposed to be only the change in width. So apparently the chord is 40 mm + 5×10^-6 m (though I'd expect it to be shortened rather than lengthened!), while the height of the arc (sagitta) is 13×10^-6 m. I still don't know whether to think of the arc length as identical to the original length, but we don't need to know that. In fact, I'll ignore the ∆l, as it is insignificant in terms of significant digits.

Using the formulas from http://mathforum.org/dr.math/faq/faq.circle.segment.html#8,

Case 8: You know c and h. Then

r = (c²+4h²)/(8h),
theta = 2 arcsin(c/[2r]),
s = r theta,
d = r - h,
K = r²[theta-sin(theta)]/2.​

and taking c = 40×10^-3 m and h = 13×10^-6 m, we can find the radius,

r = (c^2 + 4h^2)/(8h) = (1600×10^-6 + 4×169×10^-12)/(8×13×10^-6) = 15.38 m
​

The angle subtended by the arc is

theta = 2 arcsin(c/[2r]) = 2 arcsin(40×10^-3/[2×15.38]) = 0.149°
​

Check me on the calculations!

Now, to find coordinates, you can use a little trig. The angle along the arc will be proportional to the distance along the length in the unbent form (presumably), so you can work out the angle along the arc, and use sine and cosine to find the coordinates. I don't think anything you sent shows how you want those coordinates measured, so I'm not ready to do that for you.

Thank you sir! I will try it.