do i spoil everything with my green lines?
Roughly you can say the the area outside the circles are:
TOTAL AREA = 10*20 - 2*(π*5^2) = 200 - π*50; with π*5^2 being the area of one circle.
The diagonal line divides this area in two:
AREA = 100 - π*25 = 25*(4-π)
Or you can simple move areas around due to the high degree of symmetry and calculate area this way:
AREA = 10*10 - π*5^2 = 25*(4-π)
TOTAL AREA = 10*20 - 2*(π*5^2) = 200 - π*50; with π*5^2 being the area of one circle.
The diagonal line divides this area in two:
AREA = 100 - π*25 = 25*(4-π)
Or you can simple move areas around due to the high degree of symmetry and calculate area this way:
AREA = 10*10 - π*5^2 = 25*(4-π)
Attachments
I ruined everything, didn't I ....
you didn't.
see "part two" in cal weldon's post 1122: https://www.diyaudio.com/forums/the-lounge/351277-lighten-mood-113.html#post6815208
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I have had to go down the same segment road, stv, and I don't like it....
Circular segment - Wikipedia
Segment A / Circle Area = (x - sin x) / 2 Pi... where x is the angle subtended by the segment as seen from the centre of the circle. The solution is apparently irrational and transcendental in general.
The diagonal angle is horrible... 26.56505118... degrees. 0.463647609... in radians. But expressable as trig functions of roots.
Have reduced the problem to sines and Pi now. I am hoping some stuff will cancel out and I get some sort of neat algebraic solution for this particular 2 by 1 rectangle. Might be some ratios of square roots. Nearly there! I hope.
Circular segment - Wikipedia
Segment A / Circle Area = (x - sin x) / 2 Pi... where x is the angle subtended by the segment as seen from the centre of the circle. The solution is apparently irrational and transcendental in general.
The diagonal angle is horrible... 26.56505118... degrees. 0.463647609... in radians. But expressable as trig functions of roots.
Have reduced the problem to sines and Pi now. I am hoping some stuff will cancel out and I get some sort of neat algebraic solution for this particular 2 by 1 rectangle. Might be some ratios of square roots. Nearly there! I hope.
Attachments
you didn't.
see "part two" in cal weldon's post 1122: https://www.diyaudio.com/forums/the-lounge/351277-lighten-mood-113.html#post6815208
And still ....
But as I use to say: Why make anything cheap, simple and durable - when you can make it expensive, complicated and crappy???
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We need to know the area of the segment of the circle contained within the small right angled triangle at bottom left of Allen's diagram.
Area of small red bit = area of small right angled triangle - (area of circle segment + area of large red bit)
There is a formula for calculating the area of the circle segment to which I have previously given a link.
It will be now seen that the final solution hinges on the determination of the angle θ.
I don't know if this method satisfies the mathematical purists, but it works!
Area of small red bit = area of small right angled triangle - (area of circle segment + area of large red bit)
There is a formula for calculating the area of the circle segment to which I have previously given a link.
It will be now seen that the final solution hinges on the determination of the angle θ.
I don't know if this method satisfies the mathematical purists, but it works!
Attachments
How do you groom a boxer??
You've heard of self cleaning ovens? Well...
When done, there's enough to knit a pair of socks.Oh. A ghostly wool-scape.
The Mrs. always does that with his fur. She likes to freak me out with that 'ghost' thing.
Sigh, I love her anyway.
When done, there's enough to knit a pair of socks.
Or build a loudspeaker enclosure around?
And what a fine looking gentleman that Magnus is.it's my brother's brother sitting in the armchair
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