1. ## Cross Area

If the diamter of the circle is 38" and the cross beams are 6" thick, each, what percentage of the area of the circle do the beams cover? and what is the area in square inches?

2. ## Re: Cross Area

The beams cover <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">69.37%</span hide> of the circles area.
The area of the beams is <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">786.78</span hide> square inches.

3. ## Re: Cross Area

I have a different area, Rob <img src=/S/bagged.gif border=0 alt=bagged width=22 height=22>

4. ## Re: Cross Area

I've been recalculating and I see where I have made some invalid assumptions. I can now see how to solve this super puzzle but fatigue is taking hold. I shall post my solution tomorrow. No doubt others will have solved it by then
Ciao.

5. ## Re: Cross Area

I get <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">36.97%</span hide> = <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">418.098 sq. in</span hide>

6. ## Re: Cross Area

Did you make a "typo" in your percentage?

I agree with your area, but I get ".8", not ".9" in the percentage: <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">36.87% = 418.098 in

7. ## Re: Cross Area

Yes, that's a typo, thanks. The unrounded value in Excel is <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">36.8655853156343%</span hide>

8. ## Re: Cross Area

<P ID="edit" class=small>(Edited by sdckapr on 09-Dec-06 05:55. Added PS)</P> <img src=/S/yep.gif border=0 alt=yep width=15 height=15> That is what I also got, all 15 digits....

All with the simple formula:
<span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">=(4*(ASIN(Thickness/Diameter)*(Diameter/2)^2 + Thickness/2*SQRT((Diameter/2)^2-(Thickness/2)^2))-Thickness^2)/PI()/(Diameter/2)^2

=(4*(ASIN(6/38)*(38/2)^2 + 6/2*SQRT((38/2)^2-(6/2)^2))-6^2)/PI()/(38/2)^2</span hide> <img src=/S/smile.gif border=0 alt=smile width=15 height=15>

Steve

Though I guess this is a little simpler (the more it is simplified the less the "logic" can be seen...)
<span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">=(ASIN(6/38)*38^2 + 6*SQRT(38^2-6^2)-6^2)/PI()/(38/2)^2</span hide>

9. ## Re: Cross Area

<img src=/S/blackteeth.gif border=0 alt=blackteeth width=20 height=20>

10. ## Re: Cross Area

<P ID="edit" class=small>(Edited by AlanMiller on 10-Dec-06 00:20. )</P>I get <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">43.68%</span hide> = <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">396.65 sq. in.</span hide> ... but then again, I've just woken up. <img src=/S/doze.gif border=0 alt=doze width=15 height=15>

Alan

Edit - And now that I'm properly awake, I can see where I went wrong!

11. ## Re: Cross Area

<img src=/S/thumbup.gif border=0 alt=thumbup width=15 height=15>

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