MegaSack DRAW - This year's winner is user - rgwb
We will be in touch
I am trying to find an equation which equates the reduction in diameter of a circle to its reduction in area. The area of the circle before it has reduced in area is always known.
For example, a circle with a diameter of 12mm has an area of (3.14x6^2) 113mm2. if the diameter reduces by 25% to 9mm the area reduces to (3.14 x 4.5^2)63mm2.
How is this expressed as an equation, which will also work no matter what the reduction in width is?
Well the area of a circle = ? * diameter^2 / 4
So if new diameter is 75% of old diameter, new area is 0.75^2 = 0.5625 or 56.25% of original. Simply square the proportion the new diameter is of the original.
As you've said, the area is Pi*r^2. To simplify things slightly: r=d/2 ==> Area = Pi*r^2 = Pi*(1/4)*d^2. If the diameter is reduced by x percent then the new diameter will be (1-(x/100))*d (where d is still the original diameter). Now, plugging this new bit into the formula relating diameter and area gives you:
New area = Pi*(1/4)*((1-(x/100))*d)^2
=Pi*(1/4)*(d^2)*(1-(x/100))^2
= Original area * (1-(x/100))^2
Is that what you're after?
Edit: damn, beaten to it 😛
The area will change with the square of the factor by which the radius is changed (or the square of half the factor if varying the diameter).
So (assuming you're using the radius):
r = old radius
r' = new radius
A'/A = ?(r'/r)^2
D2 = D1 * sqrt((100-%)/100)
Where d2 is the smaller diameter and d1 is the original.
I'm and engineer so I think in diameters.
The application is mechanical engineering, im trying to ascertain the reduction in area (necking) of a sample once it has been subject to a tensile test without physically measuring it after the test.
The reduction is width is automatically measured using lazers.
I'm and engineer so I think in diameters.
Corner diameters? Fillet Diameters? Surely as an engineer you have to consider both.
[quote=edward2000 ]The reduction is width is automatically measured using lazers.
Is it? Or is it actually the width which is measured using lasers? What is the output of this measurement?
Not that it really matters, you just have to use the formula in a different way, just surprised that such an instrument outputs a % decrease.
Corner diameters? Fillet Diameters? Surely as an engineer you have to consider both.
Given that the original query was one that related to general geometry rather than anything too specific I don't think that any further clarification is necessary. My reason for including the comment was by way of explaining why I chose to represent the equation in diameters rather than radii that's all.
aracer - yes, the output is a percentage of the original width. I keep asking if it can measure this then why cant it measure the final diameter, but that's another argument!
[quote=edward2000 ]aracer - yes, the output is a percentage of the original width.
😯 - though that's not quite what you were suggesting before when you were talking about % reduction.
I think my first answer should work for you. Though if what you actually have is % of original diameter, then what you can do is:
% of original area = (% of original diameter)^2 / 100
The factor of 100 accounts for you using % rather than fractions, where % is 100 times the fraction.
though that's not quite what you were suggesting before when you were talking about % reduction
Agreed.
However the formulas you have given me are what im looking for!
Thanks