Originally posted by TimberWolf
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Reply to: Snowman Maths
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Previously on "Snowman Maths"
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Last year I went to the local park and tried to make the largest snowball I could manage. It was about 1.5 metres diameter before I couldn't roll it any more. Anyway I rolled other snowballs towards the first one and proceeded to amass as much snow in balls as possible. Until an old lady approached me looking annoyed and asked me what I was doing Spoilt sport.
They take some rolling when they get to about waist height, but a downward slope helps.
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Originally posted by d000hg View PostIs there a better way other than just using spades?
Far less effort and no guilt issues about the quality.
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When we had lots of snow, it was impossible to build a snowman... the snow was so dry even making a snowball was tricky, the technique of rolling it around to pick up more snow just failed.
Is there a better way other than just using spades?
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Originally posted by RichardCranium View Post
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Originally posted by TimberWolf View PostYou've calculated the surface area of a cylinder. The volume is pi . r^2 . h
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Originally posted by TimberWolf View PostAssuming a spherical body with a spherical head of half the radius of the body:
4/3.pi.r^3 + 4/3.pi.(r/2)^3 = 5 m ^3
Which comes to a radius of 1m^2 for the base and 0.5m^2 for the head, and a total height of 3m.
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Originally posted by RichardCranium View PostHmm. Were the ancient Greek philosophers that were trying to work out the equations for the square, circle, cone, cylinder, chords, diameters, pi, vectors and all that, trying to work out how to convert a rectangular garden of snow into a conical snowman with a spherical bonce?
4/3.pi.r^3 + 4/3.pi.(r/2)^3 = 5 m ^3
Which comes to a radius of 1m^2 for the base and 0.5m^2 for the head, and a total height of 3m.
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Originally posted by RichardCranium View PostEO is right.
I doubted him and I wrote an explanation and came out at the same conclusion as EO, so cancelled the post.
Now I'll write it AGAIN.
5m³ volume.
Assume cylindrical snowman (which is near enough) of 1m diameter.
Volume of cylinder = pi x d x h
5m³ = 3.142 x 1 x h
h = 5 / 3.142 = about 1.5m
EO may be daft, but he ain't stoopid.
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Hmm. Were the ancient Greek philosophers that were trying to work out the equations for the square, circle, cone, cylinder, chords, diameters, pi, vectors and all that, trying to work out how to convert a rectangular garden of snow into a conical snowman with a spherical bonce?
Leave a comment:
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Originally posted by RichardCranium View PostEO is right.
I doubted him and I wrote an explanation and came out at the same conclusion as EO, so cancelled the post.
Now I'll write it AGAIN.
5m³ volume.
Assume cylindrical snowman (which is near enough) of 1m diameter.
Volume of cylinder = pi x d x h
5m³ = 3.142 x 1 x h
h = 5 / 3.142 = about 1.5m
EO may be daft, but he ain't stoopid.
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Originally posted by TimberWolf View PostYeah, well your cylinder is 2m wide and only 1.5m tall. A very plump snowman. Mine is 5m tall.
1m
One metre.
and if you can get a 5m tall snowman out of my model, it will snap and fall over.
PL - you have the spec - get digging
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Originally posted by TimberWolf View PostYeah, well your cylinder is 2m wide and only 1.5m tall. A very plump snowman. Mine is 5m tall.
I doubted him and I wrote an explanation and came out at the same conclusion as EO, so cancelled the post.
Now I'll write it AGAIN.
5m³ volume.
Assume cylindrical snowman (which is near enough) of 1m diameter.
Volume of cylinder = pi x d x h
5m³ = 3.142 x 1 x h
h = 5 / 3.142 = about 1.5m
EO may be daft, but he ain't stoopid.
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"Waste of time, maths. You'll never use it once you leave school."
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Originally posted by EternalOptimist View PostIf you examine the source code, you will see that I have proposed a cylinder, not a sphere.
guide the cylinder into an anthropomorphic shape with spoons. agw
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