Type of curvature of gradient force from edge to center of a sphere

AI Thread Summary
The discussion revolves around identifying the type of curve generated by a set of points derived from a physics experiment involving a ball on a table. The points suggest a relationship that may be hyperbolic, parabolic, or elliptical. A participant performed a curve fitting analysis using Excel, concluding that the data closely resembles a parabola with a high R² value of 0.999997. The original poster expresses curiosity about why the gradient force translates into a nearly perfect parabolic shape. Further investigation into this phenomenon is encouraged.
Sedemichra
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I was doing some simple physics with a ball resting on a table and I made this curve

(0,0) (25, 6.8) (50, 27.51) (75, 63.4) (100, 112.34) (125, 175.7) (150, 253.3) (175, 345.4)

I was wondering if anyone could identify what sort of curve it is? I am just curious.

This is not a homework problem.
 
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Sedemichra said:
I was doing some simple physics with a ball resting on a table and I made this curve

(0,0) (25, 6.8) (50, 27.51) (75, 63.4) (100, 112.34) (125, 175.7) (150, 253.3) (175, 345.4)

I was wondering if anyone could identify what sort of curve it is? I am just curious.

This is not a homework problem.

Why can't you plot this out, post it here, and tell us what you think?

Zz.
 
I got the points off of a curve that developed due to the the apparent force of gravity pushing a sphere down on a table...I can't tell if it is hyperbolic or parabolic...or maybe a section of an ellipse(if that's even different)...I have seen the curve and I am guessing it is a parabola but I am not sure how to be certain because I could do a quadratic regression in my calculator but that wouldn't really prove anything would it?
 
Here is a picture of the curve

gYVPr66.jpg


Hyperbole, Parabola, or a section of ellipse I can't tell the difference
 
I did a curve fit on Excel
The ordered pairs you gave at the beginning lie very close to a perfect parabola:
y = 0.011329238095238*x^2 - 0.009373809523815*x - 0.037083333332703
with R^2 of 0.999997
 
Khashishi said:
I did a curve fit on Excel
The ordered pairs you gave at the beginning lie very close to a perfect parabola:
y = 0.011329238095238*x^2 - 0.009373809523815*x - 0.037083333332703
with R^2 of 0.999997

Thanks man, I wonder why the force over the gradient of the circle translates proportionally into a practically perfect parabola?

I will have to do some more investigating!

Thanks again for your reply.
 

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