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Good Evening Friends,
Put on your thinking caps tonight as the quiz is a math quiz.
A farmer had a silo that is 20' in diameter and 24' tall, also it is divided into four equal parts with a door on each outside wall and a filler cap on top.
Now for the quiz, how much volume is in each compartment? Also we all know that a bushel holds approximately 3.5 CU. FT of volume, so how many bushels of crop can be stored in each compartment?
Respectfully, :)
Each compartment will hold 538.775 bushels.
Macca
Ralph
I am one of those that grew up with metric... but here goes...
I am assumeing that it has a flat top and bottom.
Total volume = 7543 cu ft (pi X 20/2 x 20/2 X 24)
each compartment = 1886 cu ft
each compartment holds 539 bushels
Paul
About 538 bushels.
A mathematician named Hall
Had a dodecahedronal ball.
The cube of its weight
Plus his pecker times eight
Was his phone number - give him a call.
Alternative last line
Was 5/8 of 4/5 of FA
Beat me you B******s
Sorry Alex, here's a problem for you.
Using the total volume of Ralph's silo, what is the height and diameter which will use the least amount of material?
538.559 Bushels
Bushels of what?
When filling the silo you will get a conical shape under the filler point - therefore is will not be 100% full.
If the crop were canola (tiny black seeds - ask Percy Schmeiser all about it :D ) - I would go for about 520 bushels. If it were wheat (crappy red wheat) or corn, the cone would be rather steeper - so the volume would be somewhat less.
If we had some stupid idiot in the silo shovelling out the crop to the walls, the volume would be about 540 bushels less the volume of the buried idiot.
The minimum surface area is attained when the height = diaQuote:
Originally Posted by PaulS
volume = 7539.8 cu feet
PI x R squared x 2R = volume
R cubed = volume / 2 x PI
R cubed = 1200
Radius = 10.6266 feet
Dia = 21.2532 feet
Height = 21.2532 feet
Beat me again!
A spherical silo of diameter 24.33' ?
O dear Ophelia, I am ill at these numbers
I have not art to reckon my groans
Hamlet Act II Scene II
I think the key here is the four doors on the exterior wall .....
Now I am no rocket scientist but if the four doors were left open wouldn't some of those pesky bushels escape? :(
Sir Stinkalot
Sir S, you raise a good point, Ralph, a clarafication, are the doors open or closed? if they are open how high is the bottom of the door from the bottom of the silo?
Reminds me of a friend who was stripping his crop with a 40ft wide header going full bore down the paddock. Problem was, he left the header outfeed auger on when doing this. Grain goes in - grain goes out. Oops :D :DQuote:
Originally Posted by Sir Stinkalot
Good Morning Friends,
You guys are a scream and many of you were very close. Yes the doors would have to be closed in order to fill the silo and the crop is never filled to the top of the dome, but level with the top of the walls.
One thing I failed to mentioned in the quiz was that the diameter was an inscribed diameter of 20'. So PI R square, or 10 x 10 = 100 x Pi carried to the forth power of 3.1416 = 314.16 Sq Ft total. Now divide that by 4 =78.54 x 24 FT high = 1884.96 CU FT per section, divide that by the 3.5 CU FT per bushel regardless of the contents, it is still a bushel. And the final answer is 538.56 bushel per compartment.
Thank you for your support.
Respectfully, :)
These answers are all assuming that the dividing walls don't rob any of the volume because they have absolutely no thickness at all. The easiest way for the farmer to ascertain the capacity of the silo compartments would be for him to ring the manufacturer :D (or consult the product manual :D )
Mick
(BTW, bushells, barrels, roods, perches, chains, grains, hundred weight aren't you glad we changed to metric?)