oval hole drawing

[John Bertotti]

Question. I know the diameter of the hole used in a plan but I want an oval. If I figure the circumference and tie a string to match then use the two pin method for ovals would it still have the same area? I think it would but would like conformation. If the circumference is the same the area inside should remain constant. Thanks John

[sunburst]

I don't think so. I'm too rusty on my geometry to know for sure, but if you carry that logic to the extreem, (a good test for ideas in general) as the oval gets wider and wider, it eventually closes to nearly nothing. That leads me to believe you decrease the area if the circumference is the same for any shape other than a circle.

On the other hand, I don't think it would make much difference - maybe no noticable difference - in the sound of the instrument to change from a round to a slightly smaller oval.

Somebody will likely post some math or a formula soon to help you out.

[Darryl Wolfe]

I don't think so either...for the similar reasons that the area of a 4" hole is is 4x larger than a 2" hole while the circumference is only 2x

[Darryl Wolfe]

Have fun

area of an ellipse

[John Bertotti]

f5jornal, great link. I read that reply but the example used did not change the vertical height it only stretched the circle horizontally. Thereby increasing the total circumference of the item. I can not see how , regardless of shape, the area can change while the circumference remains the same. The beauty of the two pin method is having a known circumference. Thanks John

[wtaylor]

The area of an elipse is (refer to diagram below):

pi*radius1*radius2

or

pi*(axis1*axis2)/4

- so for an elipse to have the same area as a known circle, you need to product of its 2 radii (r1*r2) to equal the circle's radius squared. (or the product of its 2 axes to equal the circle's diameter squared)

Note that as you spread your 2 pins, the area of the resulting elipse becomes smaller - from the limiting case of a circle, when the interpin distance is 0, r1=r2= 1/2 your string length; as the pins spread, the major radius becomes less than this (1/2 string length minus the 1/2 the interpin distance*2), and the minor radius becomes considerably smaller.

[Tim]

reesaber -

Look at the diagram in wtaylor's message. #The two pins you are referring to would be placed close together to make that oval/ellipse. #Now imagine putting them so far apart that the circumference loop barely fits. #The r1 value in that diagram will be close to zero making the area close to zero even though the circumference is the same. That's how the area can change while the circumference changes.

[John Bertotti]

Tim I just never thought of it like that. Thanks John

Here a reply I got with the math from the link provided above. I can see why it can't work but much of this math is still hard for me to comprehend.

I take it you make a piece of string to fit along the circle so that
its length is the circumference of the circle, and then put the loop
of string around the pins and use it to draw the ellipse. This will
not produce an ellipse with the same perimeter as the circle. The
perimeter of the ellipse is not the same as the length of the string.

Suppose you could make an ellipse with the same perimeter as the
circle. Two figures with the same perimeter do not necessarily have
the same area -- in fact, they rarely do, and *no* figure with the
same perimeter as a circle will have the same area as the circle.
This is because the circle is the plane figure with minimum perimeter
for a given area, or equivalently, the plane figure with maximum area
for a given perimeter. Any other figure with the same perimeter will
necessarily have a smaller area.

If you want to get an ellipse with the same area as a circle, you
have to choose the semi-major and semi-minor axes for the ellipse so
that their geometric mean (that is, the square root of their product)
is the radius of the circle. Then the area of the ellipse will have
the same area as the circle.

In terms of the string length L and the pin separation D, I find that
the semi-major axis, a, and the semi-minor axis, b, are

##a = (L-D)/2
##b = sqrt(L^2-2LD)/2

In order to have the same area as a circle of radius R, we must have

##R^2 = ab
######= (L-D)sqrt(L^2-2LD)/4

which is quite a mess. If D is fixed, we get a quartic equation
(fourth power polynomial) in L, which is very difficult to solve. I
won't try. Suffice it to say that it is not easy to make an ellipse
with the same area as the circle using the two-pin method.

- Doctor Rick, The Math Forum