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Let's say we have a cone, and then we cut away the top of it so we get a shape similar to a bucket - how do you write the volume formula for this?
I thought it would be very easy to find out, just subtract the volume of the part you cut away from the volume of the cone as it looked in the beginning - but I always manage to get the height from the part I cut away into the equation. :confused:
I suppose trigonometry would make this easier, but I'm not sure how to use it here.
I would go along with the whole "subtract the amount you took away" routine.
I'm going to move this to the Help Forum. :)
Yeah, but the problem with that is that I want the formula to be completely independent on the part I subtract from it;Quote:
Originally Posted by Loony BoB
so that you won't have to estimate the height of that figure everytime you want to find out the volume of this bucket-figure thing. :p
Basically a formula that only uses variables found in the figure itself.
I tested if you could use the average area of the top and bottom and then calculate everything as if the figure were a cylinder, but that doesn't seem to work either, nor does the average of their radii (or that's what I imagine).
You would need to know the height/diameters of your 'bucket' and the angles of the slopes in order to calculate the volume, and as the height of the 'missing cone' can be calculated by those variables alone, you should be able to calculate everything. Nothing should ever be estimated!
there's some stupid integration, you'd integrate all the way around the circle, then integrate again along the length of whatever was left. I don't like cal 3 and can't tell you though.
r and R are the radii of the circles on the top and bottom of the figure. h is the height of the figure.
V = π(r2 +r*R + R2)h/3
Also this shape is called a Frustum. :p
I think you can derive that from calculus, yes, but it been years since I've done that :p Interestingly, that formula also works for a normal cone as well, because it's just a special case frustum with one circle with radius 0 - substituting one of the radii with 0 wil give you the well-known formula for the volume of a cone :pQuote:
Originally Posted by qwertysaur
Conical Frustum -- from Wolfram MathWorldQuote:
Originally Posted by crono_logicalI think you can derive that from calculus, yes, but it been years since I've done that :p Interestingly, that formula also works for a normal cone as well, because it's just a special case frustum with one circle with radius 0 - substituting one of the radii with 0 wil give you the well-known formula for the volume of a cone :pQuote:
Originally Posted by qwertysaur
find the height and width of the section removed, and take the volume of the section away from the total. how is this hard?
You don't know the height of the full cone though, and finding that would be a lot more work than using what you already know. :pQuote:
Originally Posted by Hypoallergenic Cactuar
use h1 and h2, h1 being the height of the entire cone, and h2 being the the second one. or how about:
(h/2 x (r/2 x r/2)pi) /3
you get the cone cut in two halves, then the radius will be half the original
Quote:
Originally Posted by Hypoallergenic Cactuar
You do not know h1 though. All you do know are the radius of the top circle, the radius of the bottom circle, and the distance between the two circles.
okay, I will call the radius of the big circle R
the little one r
and the distance between h
okay you take the sin(-1) of 1/2 the difference of R and r over hypoteuse which is the square root of that same number squared plus the height squared to get the angle of the cone on a 2 dimensional plane, then you treat that "cone" triangle as an isosceles, so you double that angle, subtract from 180, and divide by two, you now have the numbers needed to solve the equasion, that most recent number is half the angle of the 3rd angle of the triangle, and you have the distance from the first angle to the middle R, and so its just tangent of the half angle at the top of the triangle x R, now you have the full height of the cone, I assume you can do the rest.