+0  
 
0
528
1
avatar+4 

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.

eschoen  Aug 6, 2014

Best Answer 

 #1
avatar+27246 
+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
 #1
avatar+27246 
+10
Best Answer

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

20 Online Users

avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.