Making graph from coordinates when we don't know which curve it is

[amolaher750]

How did you obtain the numbers in your list? Evidently you did it by ray tracing; but how did you decide at what point along a ray the surface should lie?

And have you studied optics to find how such lenses are designed, or are you trying to do this directly from Snell's Law without using existing knowledge in the field? I know that a lot has been done about designing lenses; I also know that there are reasons that no lens can be perfect (e.g. if it focuses rays parallel to the axis, it will not exactly focus rays in other directions). I don't know much more than that.

i have researched on lens optics and want to know how the rays travels so that we can able to increase the capacity of our product that means its range, right now i am just interested how to get the curve equation so that i can back t

 

[amolaher750]

How did you obtain the numbers in your list? Evidently you did it by ray tracing; but how did you decide at what point along a ray the surface should lie?

And have you studied optics to find how such lenses are designed, or are you trying to do this directly from Snell's Law without using existing knowledge in the field? I know that a lot has been done about designing lenses; I also know that there are reasons that no lens can be perfect (e.g. if it focuses rays parallel to the axis, it will not exactly focus rays in other directions). I don't know much more than that.

because of the lens width is too big i.e. 7.14mm to use lens rule we have to go for following method:
i got that first i put i1 by myself like 0.1 and i assume the distance from light source to lens like 14 then by simple trigonometry got h1 then i used snells law to find out r1 and then calculated c using simple rule (tanc=sin(r2)/cos(r2)-(1/meu)) as c making at out is equal to normal making angle to x-axis then as for 0.1 the dx is so small so i took width of max with of lens x then from that using r1 calculated h2 so then adding h1 and h2 got y cordinate and x cordinate got by as exterior angle rule the angle between lense and x-axis is c and from c and h1 got dx and after subtracting dx from with of lens got x coordinate that are (x,y).

as i got coordinates that's why asking for how to make curve equation, i also used the cubic equation as you suggested but it satisfying only that coordinate which i took its not satisfying other remaining coordinates. so is there other way to solve or let me know how to imply cubic equation correctly to these coordinates.

 

[Dr.Peterson]

I'm sorry; I just can't follow your writing.

But you seem to be saying that the cubic equation we got (which is very accurate according to the data you provided) doesn't fit some additional data that you haven't mentioned. That suggests that what you've done is somehow not general enough.

I don't know optics enough to advise you on the overall project; I do recall that the reason we use things like "thin lens" formulas and spherical lenses is precisely that it is very difficult to work out the exact shape of an aspheric lens for a specific purpose.

 

[amolaher750]

I'm sorry; I just can't follow your writing.

But you seem to be saying that the cubic equation we got (which is very accurate according to the data you provided) doesn't fit some additional data that you haven't mentioned. That suggests that what you've done is somehow not general enough.

I don't know optics enough to advise you on the overall project; I do recall that the reason we use things like "thin lens" formulas and spherical lenses is precisely that it is very difficult to work out the exact shape of an aspheric lens for a specific purpose.

sir can you explain how to apply cubic equation in detail for data i provide?

 

[Dr.Peterson]

The first entry is x=6.999886, y=0.03258.

The equation I got from Excel after swapping the columns says y = -0.0022x^3 - 0.0353x^2 - 0.0498x + 7.0278.

This really means x = -0.0022y^3 - 0.0353y^2 - 0.0498y + 7.0278, so we evaluate -0.0022(0.03258)^3 - 0.0353(0.03258)^2 - 0.0498(0.03258) + 7.0278 = 7.0261399705081310736.

This is not accurate, presumably because the calculation is highly sensitive to rounding of the coefficients. I am not surprised by that, but had not tried doing the calculation myself. You'll have to use another method to find the coefficients, which I have to think about more. Someone else may have a ready answer.

 

[Dr.Peterson]

I had to look up how to use the LINEST function, which I haven't done much with, but once I did, it gave me much better coefficients. Here is the cubic function with appropriate precision (which still does not give exact values, but is within 1% until x gets very small):

y =-0.002249148x^3 - 0.035263315x^2 - 0.04979022x +7.027761627​

Here is one source of the information I needed: http://www.exceltoolset.com/getting-coefficients-of-chart-trendline/

 

[amolaher750]

si

I had to look up how to use the LINEST function, which I haven't done much with, but once I did, it gave me much better coefficients. Here is the cubic function with appropriate precision (which still does not give exact values, but is within 1% until x gets very small):

y =-0.002249148x^3 - 0.035263315x^2 - 0.04979022x +7.027761627​

Here is one source of the information I needed: http://www.exceltoolset.com/getting-coefficients-of-chart-trendline/

sir but when i put my values of x in that formula that you gave me it is not giving me value for y

 

[amolaher750]

I had to look up how to use the LINEST function, which I haven't done much with, but once I did, it gave me much better coefficients. Here is the cubic function with appropriate precision (which still does not give exact values, but is within 1% until x gets very small):

y =-0.002249148x^3 - 0.035263315x^2 - 0.04979022x +7.027761627​

Here is one source of the information I needed: http://www.exceltoolset.com/getting-coefficients-of-chart-trendline/

i tried this equation that you gave me but its not giving correct value of y when i put x

 

[Dr.Peterson]

I hope you recalled that the equation I gave was for the reversed data; I should have written it as x =-0.002249148y^3 - 0.035263315y^2 - 0.04979022y +7.027761627, rather than just as Excel gave it to me. It always calls the input x, whatever it really is.

But you still won't get exact values. As I said, this is not an exact equation (even though R² = 1, which means it is about as good as it gets). This is a result of curve fitting.

You could make it closer by using a higher-degree polynomial; but the fact is, your curve is not a polynomial! It's conceivable that by solving some system of equations one could find an exact formula for the ideal curve; but most likely that can't be done algebraically. This is just the way life works. As I've said, exact lens shapes are an inherently difficult problem.

 

[amolaher750]

I hope you recalled that the equation I gave was for the reversed data; I should have written it as x =-0.002249148y^3 - 0.035263315y^2 - 0.04979022y +7.027761627, rather than just as Excel gave it to me. It always calls the input x, whatever it really is.

But you still won't get exact values. As I said, this is not an exact equation (even though R² = 1, which means it is about as good as it gets). This is a result of curve fitting.

You could make it closer by using a higher-degree polynomial; but the fact is, your curve is not a polynomial! It's conceivable that by solving some system of equations one could find an exact formula for the ideal curve; but most likely that can't be done algebraically. This is just the way life works. As I've said, exact lens shapes are an inherently difficult problem.

thanks for the help sir, this is so useful.