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GEOMETRY GURUS!!! Area of an Oval - SOLVE! (pg. 2)
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| whiskers |
here's my 0.02:
find the equations for the circles (you know the radii)
guess and check where they should intersect
then place each circle on a set of axes to find the intercepts and find the area under each parth. then find the area of the rectangle.
if all fails, make a detailed cardboard model, cut it into 5 pieces like i've shown, and put teh circles over a set of coord. axes and measure where they intersect, if you can't figure out by equations.
and P.S. an oval = ellipse, and what you have is an area of 4 intersecting circles, but it's NOT an oval. |
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| jinxed84 |
| oh my god, math hurts my brain, check online maybe. i think math.com has a bunch of resources you might be able to find your answer there, or a math forum (yes there are actually forums devoted to math) good luck man |
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| mezzir |
i really think this is possible, just it would take way too much work for me
k so here's my thoughts
since you only have the radius' of the two circles, you find the derivatives of the two, and at the places where the circles intercect each other, their derivatives would = each other
so with the derivate of that point plus the radius with some weird more derivative you can find the angle of the arc that we see for each circle
using the arc length and the radius, we can find the arc length
with that we can find the area of the sector created with (the line that intersects the two points of intersetction for the larger circles) and (the arc)
so you do that for both circles, then you have the area of all but the inner rectangle, which since you know the length of the straight line from the sectors you can find it easily
anyone follow that? |
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| whiskers |
| quote: | Originally posted by mezzir
i really think this is possible, just it would take way too much work for me
k so here's my thoughts
since you only have the radius' of the two circles, you find the derivatives of the two, and at the places where the circles intercect each other, their derivatives would = each other
so with the derivate of that point plus the radius with some weird more derivative you can find the angle of the arc that we see for each circle
using the arc length and the radius, we can find the arc length
with that we can find the area of the sector created with (the line that intersects the two points of intersetction for the larger circles) and (the arc)
so you do that for both circles, then you have the area of all but the inner rectangle, which since you know the length of the straight line from the sectors you can find it easily
anyone follow that? |
the derivatives aka the slopes won't be equal at points of intersection, but the xy coordinates would be.
plus a derivative of a point won't do you much good |
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| mezzir |
| quote: | Originally posted by whiskers
the derivatives aka the slopes won't be equal at points of intersection, but the xy coordinates would be.
plus a derivative of a point won't do you much good |
hm well i'm still just taking calc so idk everything bout derivatives....but
since at the place that the two circles meet, the arc is smooth and all, wouldn't the derivatives at that point be =?
i'm talking bout where the x.y coordinates and the derivatives are =
yeah dammit this is hard
and i got all involved and now i wanna figure it out :( |
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| DJ_NRG |
| quote: | Originally posted by mezzir
yeah dammit this is hard
and i got all involved and now i wanna figure it out :( |
Seems so friggin easy, doesn't it!? That's what I told myself about 3-4 hours ago! :( |
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| Boomer187 |
| damn, I threw out my autocad 2004 disk. I will find it on a computer on campus and try to find that option I was thinking of...or just sit in front of the screen looking like a jackass. |
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| mezzir |
hm, well on second thought the derivatives part, i wasn't thinking real hard about it
and i think i have it, lmme just do a bit more algebra then i'll post it if it makes sense
lotsa geometry
:whip: :whip: :whip: |
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| mezzir |
hm...arrived at an answer but i'm unsure of it
~725.5 sq units of w/e seem reasonable? |
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| mezzir |
oh come on....anyone?
i'm just asking if its reasonable, that was the first actual math i did and like 1/2 way enjoyed and i don't even know if i did it right |
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| DigiNut |
I think they need it to be exact, not reasonable.
If you know that it's 4 intersecting circles, why can't you show us where those circles are? It says they meet at a tangent point. Great, the 3 radii shown (I'm assuming the 13.6120 is supposed to be a diameter of one of the circles) don't even have a common intersection point. What the hell?
Please clarify. If you designed this, you can surely give more info. |
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| <--ME--> |
| quote: | Originally posted by mezzir
...
anyone follow that? |
Nope! |
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