Anyone good at math, semi-circles and areas PLEASE HELP!!!!
June 26, 2013 11:58 AM Subscribe
So I am trying to figure out how to charge for a product that is a semi-circle. I charge per square feet.
The client ordered a product that was 25 feet in diameter and 6 feet deep (radius if it were a full circle), then changed it to 20 feet in diameter and 10 feet deep (radius if it were a full circle.
I tried to figure out the area of the space as half of 3.14 x radius squared. But there is no part of the equation accounting for the change in diameter... a little confused since this is the first time we are doing a semi circle.
I feel like the it shouldn't work out to the same price. If I were charging $1 per square foot, what would the difference in price be?
Well, the radius is by definition one half of the diameter, so you only need one of those figures to calculate the area (2 x r x pi) / 2 or (d x pi) / 2.
First area is 39.3 sqft, second is 31.4 sqft, so the difference in area is 7.9 sqft
posted by dabug at 12:05 PM on June 26, 2013
First area is 39.3 sqft, second is 31.4 sqft, so the difference in area is 7.9 sqft
posted by dabug at 12:05 PM on June 26, 2013
Response by poster: Ok, think of a stage that is a semi circle. The radius I'm thinking of from the back of the stage to the widest arct. Is it not?
posted by soooo at 12:05 PM on June 26, 2013
posted by soooo at 12:05 PM on June 26, 2013
Are you thinking of a circle segment? Calculation for that is here.
posted by pipeski at 12:06 PM on June 26, 2013 [1 favorite]
posted by pipeski at 12:06 PM on June 26, 2013 [1 favorite]
The client ordered a product that was 25 feet in diameter and 6 feet deep (radius if it were a full circle)
A semicircle would be half a circle, so the radius would be half the diameter. It sounds like what you are describing here is a segment. And you already quoted a price for the segment, and now want to quote a price for the semicircle, using the same $/sq ft?
posted by payoto at 12:07 PM on June 26, 2013
A semicircle would be half a circle, so the radius would be half the diameter. It sounds like what you are describing here is a segment. And you already quoted a price for the segment, and now want to quote a price for the semicircle, using the same $/sq ft?
posted by payoto at 12:07 PM on June 26, 2013
Just find for the area of the circle, then divide it in half: (π×102)/2=157sq.ft.
As for the difference, that's impossible, since the first order -- "25 feet in diameter and 6 feet deep (radius if it were a full circle)" -- is gobbledygook. Is there some reason the difference is important?
posted by Sys Rq at 12:07 PM on June 26, 2013
As for the difference, that's impossible, since the first order -- "25 feet in diameter and 6 feet deep (radius if it were a full circle)" -- is gobbledygook. Is there some reason the difference is important?
posted by Sys Rq at 12:07 PM on June 26, 2013
The only way I can make sense of the figures you've given is if you don't have a true circle here, but something that's more of an oval/ellipse. Is that the case?
Or, on preview, I see that a segment also makes sense.
posted by yuwtze at 12:15 PM on June 26, 2013
Or, on preview, I see that a segment also makes sense.
posted by yuwtze at 12:15 PM on June 26, 2013
doh ignore my dumb comment above, that's the circumference not the area. the area of the first semicircle is 245 and the 2nd is 157, difference is 88 sqft
posted by dabug at 12:16 PM on June 26, 2013
posted by dabug at 12:16 PM on June 26, 2013
The area of the segment you initially quoted would be 104.5 square feet. The area of the semicircle that the client switched to would be 157.1 square feet, so if I am understanding the scenario and question correctly you would increase the quote by 50% (multiply your original quote by 1.5).
posted by payoto at 12:17 PM on June 26, 2013 [2 favorites]
posted by payoto at 12:17 PM on June 26, 2013 [2 favorites]
The first thing you're describing sounds to me like half of an ellipse, not half of a circle. As others have pointed out, a circle has a single, constant radius, so it doesn't make sense to say that it has diameter x but radius not equal to x/2.
With an ellipse, though, you could have diameter along one direction and diameter along the other as well. Then it makes sense. (Of course, the way the thing curves might make it something other than an ellipse.)
A circle is just a special case of an ellipse where the two foci of the ellipse are not distinct -- they are the same point.
As to how to figure the cost. If the thing is a genuine semi-circle, then the cost is (1/2) * pi * radius^2 * dollars/square-foot. If the thing is a half-ellipse, then the cost should be (1/2) * pi * radius_1 * radius_2 * dollars/square-foot, where radius_1 is the shorter radius and radius_2 is the longer one. (In your first case, it would be (1/2) * pi * 12.5 * 6 * $/ft^2.)
posted by Jonathan Livengood at 12:17 PM on June 26, 2013 [1 favorite]
With an ellipse, though, you could have diameter along one direction and diameter along the other as well. Then it makes sense. (Of course, the way the thing curves might make it something other than an ellipse.)
A circle is just a special case of an ellipse where the two foci of the ellipse are not distinct -- they are the same point.
As to how to figure the cost. If the thing is a genuine semi-circle, then the cost is (1/2) * pi * radius^2 * dollars/square-foot. If the thing is a half-ellipse, then the cost should be (1/2) * pi * radius_1 * radius_2 * dollars/square-foot, where radius_1 is the shorter radius and radius_2 is the longer one. (In your first case, it would be (1/2) * pi * 12.5 * 6 * $/ft^2.)
posted by Jonathan Livengood at 12:17 PM on June 26, 2013 [1 favorite]
Yes, Payoto is correct. The first example is a segment of a circle, with an area of approximately 104sf. The second example is a segment that happens to be a true half-circle, with an area of approximately 157sf.
If you have to do these calculations regularly, I'd suggest investing in a simple drafting program, so you could just sketch the shapes and let the program tell you the area. You could even use Sketch-Up, which is free. Its area calculations aren't super accurate, but they're usually good within 1sf or so. And at $1 a square foot, you'd probably save enough time to make up for the minor inaccuracy.
posted by Kriesa at 12:25 PM on June 26, 2013
If you have to do these calculations regularly, I'd suggest investing in a simple drafting program, so you could just sketch the shapes and let the program tell you the area. You could even use Sketch-Up, which is free. Its area calculations aren't super accurate, but they're usually good within 1sf or so. And at $1 a square foot, you'd probably save enough time to make up for the minor inaccuracy.
posted by Kriesa at 12:25 PM on June 26, 2013
Best answer: D = 2r
A = pi*r^2 or A =pi(D/2)^2.
For a semi-circle, divide A by 2.
But I think this whole exercise may be at least partly unnecessary.
You're charging per square foot, but that doesn't necessarily mean that you should charge for exactly the square footage of the surface in question. Unusual shapes can involve waste. If a hardware store sells plywood in a 4'x8' sheet, and you only want a 4'x4' square, you still have to buy the whole sheet at full price. The fact that you don't need it all isn't their problem. That's the way the product comes, and if you don't want to use it all, that's your nevermind.
So depending on what you're doing here, you may be in a situation where you should actually be charging for a rectangle, where D is the long sides and r the short sides. Or maybe something between that and the semi-circle the client actually wants. You need to understand your product in addition to the math. Indeed, knowing your product really well may mean you wind up doing less math in the future. Something to think about anyway.
posted by valkyryn at 12:42 PM on June 26, 2013 [3 favorites]
A = pi*r^2 or A =pi(D/2)^2.
For a semi-circle, divide A by 2.
But I think this whole exercise may be at least partly unnecessary.
You're charging per square foot, but that doesn't necessarily mean that you should charge for exactly the square footage of the surface in question. Unusual shapes can involve waste. If a hardware store sells plywood in a 4'x8' sheet, and you only want a 4'x4' square, you still have to buy the whole sheet at full price. The fact that you don't need it all isn't their problem. That's the way the product comes, and if you don't want to use it all, that's your nevermind.
So depending on what you're doing here, you may be in a situation where you should actually be charging for a rectangle, where D is the long sides and r the short sides. Or maybe something between that and the semi-circle the client actually wants. You need to understand your product in addition to the math. Indeed, knowing your product really well may mean you wind up doing less math in the future. Something to think about anyway.
posted by valkyryn at 12:42 PM on June 26, 2013 [3 favorites]
I'm having trouble understanding the question. If it is really just a semicircle, a bunch of people above have the right answer already. But reading this part:
was 25 feet in diameter and 6 feet deep (radius if it were a full circle), then changed it to 20 feet in diameter and 10 feet deep
makes me wonder - are you talking about a semicircular annulus? That is, a large half circle, with a smaller half circle taken out of the center? If so, area of a circle = pi*R*R; area of a semicircle = 1/2*pi*R*R; area of a semicircular annulus = 1/2*pi*(R1*R1 - R2*R2).
So the way I'm reading your question, the original situation was R1=25 feet and R2 = (25-6) feet, so that the annulus was 6 feet wide; the area would have been 1/2*pi*(25*25 - 19*19).
The new situation is R1=20 feet and R2 = (20-10) feet, so the area is now 1/2*pi*(20*20 - 10*10).
IF this is what you meant, the ratio of areas is
New:Old
= (20*20-10*10)/(25*25-19*19)
= 300/264 = 1.13636.
So if you were charging $100 originally, you should charge $113.64.
However, chances are I've misunderstood what you're asking...
posted by RedOrGreen at 12:43 PM on June 26, 2013
was 25 feet in diameter and 6 feet deep (radius if it were a full circle), then changed it to 20 feet in diameter and 10 feet deep
makes me wonder - are you talking about a semicircular annulus? That is, a large half circle, with a smaller half circle taken out of the center? If so, area of a circle = pi*R*R; area of a semicircle = 1/2*pi*R*R; area of a semicircular annulus = 1/2*pi*(R1*R1 - R2*R2).
So the way I'm reading your question, the original situation was R1=25 feet and R2 = (25-6) feet, so that the annulus was 6 feet wide; the area would have been 1/2*pi*(25*25 - 19*19).
The new situation is R1=20 feet and R2 = (20-10) feet, so the area is now 1/2*pi*(20*20 - 10*10).
IF this is what you meant, the ratio of areas is
New:Old
= (20*20-10*10)/(25*25-19*19)
= 300/264 = 1.13636.
So if you were charging $100 originally, you should charge $113.64.
However, chances are I've misunderstood what you're asking...
posted by RedOrGreen at 12:43 PM on June 26, 2013
Best answer: Is it possible that you are describing the first area in terms of the actual width of the area in question and the actual depth of the area in question? You could be describing a section of a circle (less than a semi-circle) in those terms, which is not being understood by people here.
posted by ssg at 12:46 PM on June 26, 2013
posted by ssg at 12:46 PM on June 26, 2013
Response by poster: Hi SSG, yes this is what I am describing. How would I calculate the area of the space having the width and depth of the circle?
posted by soooo at 1:27 PM on June 26, 2013
posted by soooo at 1:27 PM on June 26, 2013
So like a solid cylinder?
posted by Think_Long at 1:38 PM on June 26, 2013
posted by Think_Long at 1:38 PM on June 26, 2013
Could you maybe draw a diagram for us, with labels, for the before and after situations? Even a photograph of a sketch on a napkin would be useful to clarify what you mean...
posted by RedOrGreen at 1:39 PM on June 26, 2013
posted by RedOrGreen at 1:39 PM on June 26, 2013
A picture would really help. I feel like from your description it's this, though, right?
posted by RustyBrooks at 1:52 PM on June 26, 2013
posted by RustyBrooks at 1:52 PM on June 26, 2013
A quick construction in Geogebra gives the area as 104.47 square units.
posted by Wulfhere at 1:57 PM on June 26, 2013 [1 favorite]
posted by Wulfhere at 1:57 PM on June 26, 2013 [1 favorite]
As for the second one (revised customer request,) if it is 20' wide and 10' deep it really IS a semicircle, and thus has area (1/2)*(3.14)*10^2 = 314/2 = 157 square feet.
posted by Wulfhere at 2:00 PM on June 26, 2013
posted by Wulfhere at 2:00 PM on June 26, 2013
I think this is answered now (50% more, as indicated above). For reference, I think the confusing part was 6 feet deep (radius if it were a full circle), when you actually mean a circle segment with a depth of 6 feet and a width of 25 feet. The actual radius of the circle this is a segment of is just slightly more than 16 feet.
posted by ssg at 2:02 PM on June 26, 2013 [1 favorite]
posted by ssg at 2:02 PM on June 26, 2013 [1 favorite]
Gonna make an assumption that your width is the length of the chord and your depth is the distance from the center of the chord to the arc. Pipeski's link has the calc, but you're missing the right values. You don't know R (radius) or the C (central angle). Using your names Width (W) and Depth (D)
R = (D² + 1/4W²)/2D
and
C = 2(arcsin(W/2R))
You can plug R and C into the equation in the referenced link, 'cause it's gonna look ugly trying to text format it.
posted by hwyengr at 2:03 PM on June 26, 2013
R = (D² + 1/4W²)/2D
and
C = 2(arcsin(W/2R))
You can plug R and C into the equation in the referenced link, 'cause it's gonna look ugly trying to text format it.
posted by hwyengr at 2:03 PM on June 26, 2013
Dude, you have to draw a picture. And post it. Use Google Drive's draw program if you have to.
posted by amtho at 8:55 PM on June 26, 2013
posted by amtho at 8:55 PM on June 26, 2013
This thread is closed to new comments.
posted by Kriesa at 12:04 PM on June 26, 2013