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Reply to: Snowman Maths

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Previously on "Snowman Maths"

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  • RichardCranium
    replied
    Originally posted by TimberWolf View Post
    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.
    What a Meddlesome Ratbag.

    Leave a comment:


  • TimberWolf
    replied
    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.

    Leave a comment:


  • RichardCranium
    replied
    Originally posted by d000hg View Post
    Is there a better way other than just using spades?
    Go back indoors, sit down with a beer, and let the children do it.

    Far less effort and no guilt issues about the quality.

    Leave a comment:


  • d000hg
    replied
    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?

    Leave a comment:


  • TimberWolf
    replied
    Originally posted by RichardCranium View Post
    Actually, I thought later you'd mixed radius and diameter (as EO may have done).

    Leave a comment:


  • RichardCranium
    replied
    Originally posted by TimberWolf View Post
    You've calculated the surface area of a cylinder. The volume is pi . r^2 . h

    Leave a comment:


  • EternalOptimist
    replied
    Originally posted by TimberWolf View Post
    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

    Leave a comment:


  • TimberWolf
    replied
    Originally posted by RichardCranium View Post
    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.

    Leave a comment:


  • TimberWolf
    replied
    Originally posted by RichardCranium View Post
    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

    Leave a comment:


  • RichardCranium
    replied
    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:


  • EternalOptimist
    replied
    Originally posted by RichardCranium View Post
    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.

    Leave a comment:


  • EternalOptimist
    replied
    Originally posted by TimberWolf View Post
    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



    Leave a comment:


  • RichardCranium
    replied
    Originally posted by TimberWolf View Post
    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.

    Leave a comment:


  • RichardCranium
    replied
    "Waste of time, maths. You'll never use it once you leave school."

    Leave a comment:


  • TimberWolf
    replied
    Originally posted by EternalOptimist View Post
    If 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




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

    Leave a comment:

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