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# I have a bale of hay that is 5' in diameter and 4' wide. If I store the bale outside, the outer 8 inches gets damaged by sun and rain. What

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I have a bale of hay that is 5' in diameter and 4' wide. If I store the bale outside, the outer 8 inches gets damaged by sun and rain. What is the volume of hay that is damaged? Answer in cubic feet and as a percentage of the bale.

Aug 6, 2014

#1
+32549
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The volume of the full bale is V = w*pi*d2/4  where w is width and d is diameter.

If damage depth is s then the undamaged inner cylinder has volume Vu = (w-2s)*pi*(d-2s)2/4 because the width reduces by s at both left and right, and the diameter reduces by s on "both" sides.

The damaged volume is therefore  the difference between these. Here, we have w = 4', d = 5' and s = (2/3)' (because 8" is 2/3 of a foot). So:

$${\mathtt{V}} = {\frac{{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{2}}}}{{\mathtt{4}}}} \Rightarrow {\mathtt{V}} = {\mathtt{78.539\: \!816\: \!339\: \!744\: \!831}}$$ cubic ft.

$${\mathtt{Vu}} = {\frac{\left({\mathtt{4}}{\mathtt{\,-\,}}{\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{2}}}{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({\mathtt{5}}{\mathtt{\,-\,}}{\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{2}}}{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}}{{\mathtt{4}}}} \Rightarrow {\mathtt{Vu}} = {\mathtt{28.157\: \!978\: \!598\: \!841\: \!850\: \!5}}$$ cubic ft.

Damaged volume Vd ≈ 78.54 - 28.16 = 50.38 cubic ft.

As a percentage Vd/V ≈ 100*50.38/78.54 = 64.15%

Aug 6, 2014

#1
+32549
+10

The volume of the full bale is V = w*pi*d2/4  where w is width and d is diameter.

If damage depth is s then the undamaged inner cylinder has volume Vu = (w-2s)*pi*(d-2s)2/4 because the width reduces by s at both left and right, and the diameter reduces by s on "both" sides.

The damaged volume is therefore  the difference between these. Here, we have w = 4', d = 5' and s = (2/3)' (because 8" is 2/3 of a foot). So:

$${\mathtt{V}} = {\frac{{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{2}}}}{{\mathtt{4}}}} \Rightarrow {\mathtt{V}} = {\mathtt{78.539\: \!816\: \!339\: \!744\: \!831}}$$ cubic ft.

$${\mathtt{Vu}} = {\frac{\left({\mathtt{4}}{\mathtt{\,-\,}}{\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{2}}}{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({\mathtt{5}}{\mathtt{\,-\,}}{\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{2}}}{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}}{{\mathtt{4}}}} \Rightarrow {\mathtt{Vu}} = {\mathtt{28.157\: \!978\: \!598\: \!841\: \!850\: \!5}}$$ cubic ft.

Damaged volume Vd ≈ 78.54 - 28.16 = 50.38 cubic ft.

As a percentage Vd/V ≈ 100*50.38/78.54 = 64.15%

Alan Aug 6, 2014