- #1

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The data:

(20, 59)

(30, 44)

(40, 34)

(50, 28)

(60, 24)

(70, 21)

(80, 19)

(90, 17)

(100, 15)

(125, 12)

(150, 10)

(175, 9)

(200, 8)

(225, 7)

(250, 6)

I suppose I could add (0,infinity) to that list. Nothing above x=250 is relevant.

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- Thread starter gnome
- Start date

- #1

- 1,036

- 1

The data:

(20, 59)

(30, 44)

(40, 34)

(50, 28)

(60, 24)

(70, 21)

(80, 19)

(90, 17)

(100, 15)

(125, 12)

(150, 10)

(175, 9)

(200, 8)

(225, 7)

(250, 6)

I suppose I could add (0,infinity) to that list. Nothing above x=250 is relevant.

- #2

Integral

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- #3

uart

Science Advisor

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I have a set of data points relating the width of an object in an image to its distance from the camera. I'd like to find the simplest curve that fits "pretty well". When I graph the points, it looks like a hyperbola would be a good fit. Is there a simple iterative method to find an equation?

Yes theoretically it should be a hyperbola. So take the reciprocal of the second data column and then it should be a straight line.

- #4

matt grime

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log log plots. as were taught decades ago, but seemingly not anymore....

- #5

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Thanks, uart, that was very helpful.

Matt: I'll put that on my to-do list. ;)

Matt: I'll put that on my to-do list. ;)

- #6

matt grime

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If you believe that data x_i and y_i are related by something like x^n=k*y^m, then taking logs nlog(x)=log(k)+mlog(y), i.e. their logs should form a straight line graph. You can also try variations if you thought that y^n=k*exp(x), or something similar. You used to be able to buy log-log graph paper to do this. So I'm told - I'm too young to have used this.

- #7

Gib Z

Homework Helper

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I quite like [itex]y = 957.83 x^{-0.9057}[/itex] thank you very much :)

- #8

uart

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I quite like [itex]y = 957.83 x^{-0.9057}[/itex] thank you very much :)

Or

[tex]y = \frac{1000}{0.63 x + 3.65}[/tex]

It depends on what model you choose to fit.

- #9

Integral

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[tex] \frac 1 x + \frac 1 y = \frac 1 f [/tex]

I would guess this relationship since I know about the thin lens formula. That is the trouble with simply fitting data with no thought of the known physical relationships. You can get perfectly good fits which have no physical meaning.

- #10

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log log plots. as were taught decades ago, but seemingly not anymore....

Actually we just did them in my physics I highschool course to show Kepler's 3rd using the orbital radius and period of the planets :rofl:

But yeah the slope of the log-log graph is the power of the function.

Edit: I got:

[tex]y=\frac{957.83}{x^{.90499}}[/tex]

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