Snowman Maths

[RichardCranium]

Yeah, well your cylinder is 2m wide and only 1.5m tall. A very plump snowman. Mine is 5m tall.

EO 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.

[EternalOptimist]

Yeah, well your cylinder is 2m wide and only 1.5m tall. A very plump snowman. Mine is 5m tall.

it specifically says 1m wide.

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

[EternalOptimist]

EO 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.

[RichardCranium]

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?

[TimberWolf]

EO 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.

You've calculated the surface area of a cylinder. The volume is pi . r^2 . h

[TimberWolf]

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?

Assuming 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.

[EternalOptimist]

Assuming 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.

the only way to test the models is with a field trial. come on PL , hows it going

[RichardCranium]

You've calculated the surface area of a cylinder. The volume is pi . r^2 . h

[TimberWolf]

Actually, I thought later you'd mixed radius and diameter (as EO may have done).

[d000hg]

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?