OT-Figuring The Area of Small Circle

GarryinNC

Well-known Member
I am working on a Hesston baler that has a floating orifice in the tailgate hydraulics. You can use a .960" or an .080" orifice. I am trying to figure how much more (in percent) the larger one will flow.

I know that the formula for the area of a circle is pi times the square of the radius. What is confusing me is that usually when you square a number, you end up with more; i.e.- 4 x 4= 16. A larger number than you started with. When you square a decimal(the radius of .080 is .040), you end up with a number that is less than what you started with. .040x.040=.0016.

Is this going to work in figuring the area of the .080" circle? .0016x3.1416=.005026 sq. inch. Is that the area of the .080" circle?

Thanks,
Decimally challenged, Garry
 
yes the math works the same regardless. Just a little harder to keep the decimals in the right place when working with
very small numbers.
 
yup, that's right. Think of it this way. Squaring .50 is .5 x.50 = .250 (smaller than the original)...but think of it another way: if you take 1/2 (.5) of a half what do you have? A quarter (.250. )
 
Yes, your math is correct A = pi x .040 x .040 = .oo5 in^2. The .096 orifice will flow more in proportion to it's cross sectional area.
 
(quoted from post at 17:08:51 10/22/15) . You can use a .960" or an .080" orifice. I am trying to figure how much more (in percent) the larger one will flow.

Decimally challenged, Garry

There's a really big difference between .960 and .080.
Typo?
 
Yes, sorry, a typo, misplaced decimal. Should be .096". Would it flow about 140% of the .080" orifice.

I tried the math on a 1 inch circle and came up with .785 sq inch, which sounds about right if you visualize a 1 inch square with the corners rounded off. .5x.5=.25x3.1416=.7854.

Thanks for the replies, everyone.

Garry
 
.096 x .096/(.08 x .08) pi isn't needed

to put it in your calculator

.096 x .096 / .08 / .08 =

0r .012 x .012 / (.01 x .01)

or 1.2 x 1.2 / (1 x 1)

= 1.44 times larger
 
You're getting hung up on the decimal point and ignoring the UNITS. Area is measured in SQUARE INCHES, not INCHES. There is no "smaller" or "bigger". To say that .0016 square inches is "smaller" than .04 inches is like saying "five pounds is smaller than 52 seconds". You're trying to compare different things.
 
Here you go.

0.080 / 2 = 0.040 X 0.040 = 0.0016 X 3.1416 = 0.005027 SqIn
0.096 / 2 = 0.048 X 0.048 = 0.002304 X 3.1416 = 0.007238 SqIn

0.007238 - 0.005027 = 0.002211 Amount in SqIn that the .096 is larger than the .080

Therefore: the .096 is 42% larger than the .080
Therefore: the .080 is 30.5% smaller than the .096

Therefore: if you had the small one .080 in the baler now and changed to larger .096 you would get 42% increase.
Therefore: if you had the larger one .096 in the baler now and changed to the smaller .080 you get a 30.5% reduction.

Tom
 
I like your example. Very well written and explained. I used to sit here at night and figure how much a few thousands of an inch of piston protrusion or valve recess would affect compression ratios. Now days, I have found real good sites on the internet that takes the heavy work (using a calculator) out of it.
 
MarkB_MI, You, truly, know quite a lot about most everything, but I don't think you read my post very well. I was not comparing square inches to inches. I do know the difference. .0016 was the square of the radius, not square inches.

Thanks to all of you for taking time to calculate and post. Now if I can just just get that 5 pounds to 52 seconds comparison figured out!

And I did get the baler problem fixed. Made a .080 orifice and it keeps the slack out of the belts when tailgate is closing.

Garry
 
>I don't think you read my post very well. I was not comparing square inches to inches. I do know the difference. .0016 was the square of the radius, not square inches.

If the radius is measured in inches, then square of the radius has units of square inches. The fact that you have to multiply the square of the radius times pi to get the actual area is irrelevant, because pi is unitless.

When calculating any algebraic formula, it's necessary to include the units in the calculation. This is a point missed in algebra and geometry textbooks, but it's absolutely critical in engineering and science. And it's something we do every day without thinking about it, when we calculate bushels per acre or miles per gallon, etc.
 

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