Calculating width of parabolic reflector sheets

[cookindaddy]

I am using the formula for a parabola: y=x^2/4f and varying the focus f.

I want to calculate how far along the arms of the parabola would be covered with a thin reflective sheet of a given width bent to conform to the given parabolic shape. This is in aid to build a parabolic heater (a parabolic trough concentrator, like this one http://www.ffwdm.com/solar/images/reflector-assembled.jpg.

Since sheet material is sold in standard sizes, it is more economic to not have wasted material from cutting so 1/3 sheet, 1/2 sheet, full sheet would be three of the widths that might cover the parabola. How far along a particular parabola would that size sheet cover so that the ends of the arms may be accurately located for the design.

I have been doing this manually but there must be a way to calculate the distance along a parabolic curve.

Thank you,

George Plhak

 

[pappus]

I am using the formula for a parabola: y=x^2/4f and varying the focus f.

I want to calculate how far along the arms of the parabola would be covered with a thin reflective sheet of a given width bent to conform to the given parabolic shape. This is in aid to build a parabolic heater (a parabolic trough concentrator, like this one http://www.ffwdm.com/solar/images/reflector-assembled.jpg.

Since sheet material is sold in standard sizes, it is more economic to not have wasted material from cutting so 1/3 sheet, 1/2 sheet, full sheet would be three of the widths that might cover the parabola. How far along a particular parabola would that size sheet cover so that the ends of the arms may be accurately located for the design.

I have been doing this manually but there must be a way to calculate the distance along a parabolic curve.

Thank you,

George Plhak

1. If the curve is given by y = f(x) then the length of the curve is calculated by:

\(\displaystyle l = \int_a^b\left( \sqrt{1+\left(\frac{df}{dx} \right)^2} \right) dx\)

2. I assume that the "arms" of the parabola have equal length. Then the borders of the integral are -a and a and the length of the curve is consequently:

\(\displaystyle \displaystyle{l = \int_{-a}^a\left( \sqrt{1+\left(\frac{x}{2f} \right)^2} \right) dx = \frac{2f^2 \cdot \ln\left(\frac1{4f^2} \left(\sqrt{a^2+4f^2} +a \right)^2 \right)+ a\sqrt{a^2+4f^2}}{2 |f|}}\)

Remark: The last result was calculated by my computer, so don't try to do it by hand

 

[cookindaddy]

pappus Thanks!

Your computer told you that? Math certainly has come a long way!

Thinking a bit more about it and looking at your formula, I want to check that I understand correctly: a is the value of x that shows the horizontal position of a sheet edge of width "l" formed to match the given parabola, like this:



If so, and I know the value of "l" I suppose that what I really want to do is solve for "a".

Is it possible for you to re-arrange the formula you gave to solve for a? What software do you use to do this?

Thanks!

George

 

[pappus]

pappus Thanks!

Your computer told you that? Math certainly has come a long way!

Thinking a bit more about it and looking at your formula, I want to check that I understand correctly: a is the value of x that shows the horizontal position of a sheet edge of width "l" formed to match the given parabola, like this:

View attachment 1935

If so, and I know the value of "l" I suppose that what I really want to do is solve for "a".

Is it possible for you to re-arrange the formula you gave to solve for a? What software do you use to do this?

Thanks!

George

1. All your considerations are OK.

2. I've attached what my computer "told" me. (Note that I calculated half of l , therefore the lower border of the integral is zero!)

3. If you know already the values of l, f then a numerical value of a could be obtained quite easily.

4. I still use derive 6.0 even though TI doesn't maintain this program any more.

 

[cookindaddy]

Thank you pappus.

What I see, even enlarged is very small. Could it be enlarged somewhat? I intend to try it and I don't want to make a mistake.

Thanks for the clarifications. TI is Texas Instruments?

George

 

[pappus]

Thank you pappus.

What I see, even enlarged is very small. <--- sorry for the inconvenience
Could it be enlarged somewhat? I intend to try it and I don't want to make a mistake.

Thanks for the clarifications. TI is Texas Instruments? <--- yes

George

1. Maybe you've noticed that the computer gave 4 solutions. I've copied them to test them.

2. I used f = 25 cm and l = 60 cm ( but not sure if these values are realistic enough?) and got 2 complexe values for a which can't be real (sorry, pun intended), one negative value for a which probably doesn't work properly and one real positive value.

This is:

\(\displaystyle a = \sqrt{2} \cdot \sqrt{f\cdot \left( \sqrt{f^2 + l^2} - f \right)}\)

btw: SIGN(f) = (+1)

3. With this formula I get \(\displaystyle a \approx 44.72\) which can't be correct with a length l = 60. So maybe the length of the arms is 22.36.

4. If I were you I would draw the cross-section of the solar oven, use a piece of wire whose length correspond to the width of the reflecting sheet and test if the formula works properly. If it works you can use the formula - if it doesn't I have to apologize.

 

[cookindaddy]

I have more primitive tools. I do not get the same answer.

Here I have translated your formula to a spreadsheet:

I've used your parameters and get 425.7cm which makes no sense.

Can anyone suggest what I have done wrong?

Thanks.

George

 

[cookindaddy]

Oops - missing a bracket. Now I get the same as you.

I will try to plot the curve and check the prediction, maybe later today.

Thanks for all your help.

George

 

[cookindaddy]

I will try to plot the curve and check the prediction, maybe later today.

I made a full size mockup using a 25cm focal length parabola and a 60cm full sheet size (30cm for 1/2 of the parabola) and checked the formula prediction (not good I'm afraid) using a piece of wire 30cm long matched to the parabola outline. I've tried to be accurate to about a millimeter. The formula predicts 22.36cm but the actual result is about 28.75cm



George Plhak

 

[pappus]

I made a full size mockup using a 25cm focal length parabola and a 60cm full sheet size (30cm for 1/2 of the parabola) and checked the formula prediction (not good I'm afraid) using a piece of wire 30cm long matched to the parabola outline. I've tried to be accurate to about a millimeter. The formula predicts 22.36cm but the actual result is about 28.75cm

View attachment 1943

George Plhak

Thanks for this reply. You've done a great job - and obviously I've reached the limits of my math

When I worked in school ( long time ago) we used a circular cross-section to build a (cylindrical) solar-oven. The focal length is approximately \(\displaystyle f = \frac12 r\)

With your values:

\(\displaystyle f = 25 cm ~\implies~r = 50 cm\)

Central angle corresponding to a length of \(\displaystyle l = 60 cm\):

\(\displaystyle \displaystyle{\frac{60}{2 \pi \cdot 50}=\frac{\alpha}{360^\circ}~\implies~\alpha \approx 68.755^\circ}\)

This will give you a width of the reflector sheet of

\(\displaystyle \displaystyle{w = r \cdot \sin(\frac12 \alpha)~\implies~w \approx 28.23\ cm}\)

So it seems to me that your result is quite exact. Congratulations!

 

[pappus]

I made a full size mockup using a 25cm focal length parabola and a 60cm full sheet size (30cm for 1/2 of the parabola) and checked the formula prediction (not good I'm afraid) using a piece of wire 30cm long matched to the parabola outline. I've tried to be accurate to about a millimeter. The formula predicts 22.36cm but the actual result is about 28.75cm

View attachment 1943

George Plhak

It's me again.

I used the values to simplify the more genral result of my previous posts. And suddenly my computer works fine ...

 

[cookindaddy]

Computer glich here Pappas which caused me not to see your last message. So please ignore what I posted below which I've left in place because of the background info. I will study the revised program thanks! George


Hello Pappas and thanks.

I always knew that I could make a model and find a measurement but I still want to be able to calculate the intersection of the sheet with the parabolic curve. It seems to me that a circular reflector is not the same as a parabola and although the calculation comes close with f=25, will it match as closely at other values? My current design uses f=10cm for example.

My intent is to try a series of curves to optimize the weight balance of the whole solar reflector assembly. If it is balanced, less energy is required to move it around to follow the sun hence the system is more efficient. The motor can be very small.

So the calculation of the intercept is part of the calculation of the weight distribution.

Does anyone else have an idea how this intercept might be calculated?

Many thanks

George Plhak
some of my solar work

 

[cookindaddy]

Pappas,

Can I please ask you to turn the formula inside out, so that I am solving for a? It would be a lot more useful to me here. My computer does not know how to get from step #3 to step #4. Thanks. George

 

[pappus]

Pappas,

Can I please ask you to turn the formula inside out, so that I am solving for a? It would be a lot more useful to me here. My computer does not know how to get from step #3 to step #4. Thanks. George

I think it would be best for you to get a spread-sheet where you can vary the values of focal length and pick the appropriate value for the width of the reflector arms and the length of the reflective sheet.

If it isn't possible to attach an Excel-file here I'll be going to pm you.

As I've imagined I can't upload the spreadsheet.

So I'll try if it is possible to send you the file by pm.