Help figuring area of oddball shape
- Thread starter Dozerhand
- Start date
Bill C
Well-known member
The area of a spherical cap is given by the equation pi*(a^2 + h^2). Where “a” is the radius of the spherical cap, and “h” is the height of the cap segment.
So in your case it’s Area = pi*(24^2 + 12^2) = 3.14(576+144) = 2260.8 in^2 = 15.7 sq ft.
So in your case it’s Area = pi*(24^2 + 12^2) = 3.14(576+144) = 2260.8 in^2 = 15.7 sq ft.
Last edited:
Bobioz1
Well-known member
I concur.
https://keisan.casio.com/exec/system/1514268767
Plugging your numbers into the above site gives 402 sq in or 2.79 sq ft. The area of a circle is 3.14 x r^2. Your 5' dia is 19.62 sq ft. I'm having a problem with the above that says more than half that area is covered..
lg
no neat sig line
Plugging your numbers into the above site gives 402 sq in or 2.79 sq ft. The area of a circle is 3.14 x r^2. Your 5' dia is 19.62 sq ft. I'm having a problem with the above that says more than half that area is covered..
lg
no neat sig line
mrobins297aaa
Well-known member
I agree with Larry the area of the arc and chord in the op photo is 2.79 sq. ft
Bill C
Well-known member
It’s because we are both assuming something different regarding the question.
You are assuming it’s 2-dimensional flat profile that is semicircular in shape. In which case the 2.8sqft calc is correct. I assumed he was making some type of domed cap (spherical) defined by a 48in diameter and a dome/cap height of 12in
mizzoutrover
Well-known member
per the OP, he is cutting a plate
I agree with Larry
I agree with Larry
mrobins297aaa
Well-known member
I'm not really sure what he is asking, but in your eq isn't the radius of the of the circle 30"...........48" is not the dia.
well maybe it is, if were looking at a 48" sper dome over a 60" pond.............like looking down (a plan view).........then the difference would be the area of a 60" circle - the area 48" circle
well maybe it is, if were looking at a 48" sper dome over a 60" pond.............like looking down (a plan view).........then the difference would be the area of a 60" circle - the area 48" circle
Last edited:
mrobins297aaa
Well-known member
he's going to lose 12.56 sq. ft
Last edited:
Bill C
Well-known member
I assumed he was trying to figure out how much “screen” material he needed to cover a dome shape. I probably immediately overcomplicated the problem as simple 2d areas/projections seemed too straight forward.
As for the 48” vs 60” diameter. I figured something was a typo. He said 5ft diameter pond, but then was making a 48” diameter circle........ ANYWAY..... the more I re-read it, I think the OP is making a semi-circular flat cover that will be exactly as drawn in his picture.
Last edited:
mrobins297aaa
Well-known member
what if the 60" Id is a 60" dia. horizontal over flow pipe and the chord/arc plate is attached to one end of the 60" pipe then the correct answer would be that he is losing 2.79 sq. ft of the total area of 19.65 sq ft
Bill C
Well-known member
Which is exactly as LG figured, and the most likely answer to the OP’s somewhat ambiguous question.
mrobins297aaa
Well-known member
hahaha...maybe the real question was to figure out what the real question is
bugnut
ALLIANCE MEMBER
I tend to agree with you. My wife was a teacher and writing test questions was part of the job. A lot of times she would have me review then so there was not any ambiguity in the question that could lead to this problem or where multiple answers could be correct.
lg
no neat sig line
2ndGearRubber
Well-known member
Make 2 triangles with the 90 degree angles in the center of your circle. Add 25% to the area of those two to cover the curved sections s and waste.