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Topic: Ladder into Tower (Read 2885 times) |
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TonyMo
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Ladder into Tower
« on: Sep 8th, 2002, 9:09am » |
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Can't see any thread about this one yet, so here goes....
I'm assuming that "cylindrical" just means that the ladder has to go straight through the doorway and not at a sideways angle.
I haven't finished solving it, but here is how far I have got.
Lets assume first that the doorway is as high as necessary, and that when the ladder is at angle x from horizontal, it is at height h at the point it passes through the door.
So the far end of the ladder is at height L*sin(x) up the wall. The increase in height from the doorway to the far wall is D*tan(x). So that makes h equal to L*sin(x) - D*tan(x)
To know how high the doorway needs to be, we want to find the maximum value of h. This will be the point at which dh/dx is zero.
Differentiating the above formula we get dh/dx = L*cos(x) - D*(sec(x))^2
So when dh/dx is zero we get L*cos(x) = D/(cos(x))^2
This simplifies to D/L = (cos(x))^3, or in otherwords, cos(x) equals the cube root of D/L, where x is the angle of maximum height in the doorway.
I have verified this answer empirically in Excel.
Now I need to substitute for x back into the original formula, to get the doorway height required to accommodate a ladder of length L with tower diameter D. That is a horrendous substitution, which I have not managed to simplify yet. This is as far as I have got.
This all seems quite long-winded. Have I missed a simpler way?
Cheers,
Tony
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| « Last Edit: Sep 28th, 2002, 1:43am by TonyMo » |
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Kozo Morimoto
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Re: Ladder into Tower
« Reply #1 on: Sep 10th, 2002, 1:54am » |
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Good work on the initial steps - it got me thinking...
I've went ahead with your work and put x back in the formula etc etc and rearranged the formula in Mathematica so that I got:
L = (H^(2/3) + D^(2/3))^(3/2)
it looks like it'll simplify further...
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TonyMo
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Re: Ladder into Tower
« Reply #2 on: Sep 10th, 2002, 1:42pm » |
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Kozo- thanks for following up. Once I got my brain into gear again, I managed to substitute and simplify by hand, and got the same answer you did:
L2/3 = H2/3 + D2/3
I don't think it does simplify any further. It has a nice pythagorean look to it, based on the cube roots of the dimensions.
It's interesting to note that the formula is symmetrical with respect to the diameter of the tower and the height of the door, as the point of constraint is effectively a point: the top edge of the door.
It would be nice to hear from the originator of the problem!
BTW, William - I think this puzzle deserves an 'M' rating! 
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| « Last Edit: Sep 10th, 2002, 1:45pm by TonyMo » |
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JPM
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In light of the above use of more complicated math than I'm about to display, this may be far too amateur-ish a whack at this problem for your tastes. But being one who has spent a fair amount of time fitting ladders in and out of doorways I will claim a certain amount of experience with the problem. So if I've got a ladder in a cylindrical building and I want to get it out: if I can get it at a 45 deg angle (with part of it sticking out the door) then I can get it out the door. And if I can get it out, then I can get it back in. The longest ladder possible would sit at a 45 deg angle with one end touching the far wall opposite the door, one end touching the ground outside the door, and somewhere in the middle touching the top of the doorway. This creates a triangle with X and Y being the far wall of the building and the ground, and Z being the ladder. Since X^2 + Y^2 = Z^2 we get L^2 = (D+H)^2 + (D+H)^2. This simplifies to L = sqrt(2)(D+H).
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TonyMo
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Re: Ladder into Tower
« Reply #4 on: Sep 13th, 2002, 6:05am » |
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That was what I thought at first, until I analyzed it.
Say the diameter of the tower is 8m and the height of the door is 2m. If a 45-degree ladder touches the top of the doorway, that makes the length of the ladder sqrt(200) = 14.14m. You won't get that through the door! The maximum length of ladder that would go is 13.2m long, and would touch the top of the door when its angle was 32.2 degrees.
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Kozo Morimoto
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Re: Ladder into Tower
« Reply #5 on: Sep 27th, 2002, 5:18pm » |
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Does the question mark on the topic thread mean TonyMo haven't got the right answer yet?
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TonyMo
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Re: Ladder into Tower
« Reply #6 on: Sep 28th, 2002, 1:45am » |
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I think we are at the right answer now. The ? icon was just what I selected for my original posting. I've changed it now!
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Andrew
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If you ignore the ladders thickness you can fit an infininte ladder in, as long as its a rope ladder. 
I think the answer might be something along these lines considering:
"Note: While the contributor was serving in the German military, a superior rank tried to outsmart him with this riddle. It didn't work "
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Lijesh Shetty
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why the solution is made so complicated
why is not the simple phytogaras theorem applied directly.
if the block is cylinder, the door is in right angle with the base.
that means
(length of Ladder)=sqrt(L*L + D*D)
so the length of the ladder has to be a bit shorter then the max length.... 
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Lijesh Shetty
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why the solution is made so complicated
why is not the simple phytogaras theorem applied directly.
if the block is cylinder, the door is in right angle with the base.
that means
(max length of Ladder)=sqrt(L*L + D*D)
so the length of the ladder has to be a bit shorter then the max length....
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aero_guy
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Re: Ladder into Tower
« Reply #10 on: Mar 17th, 2003, 9:24am » |
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You don't quite understand the question. First, you don't have L, that is what you are trying to find. Second, the base isn't D, as the ladder at the point where it would get stuck would be forming a triangle where the top is against the back wall of the tower while the bottom is still outside the tower and at some point along its length is touching the top of the doorway.
Try solving the riddle with this info and you may find it a bit more difficult.
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rmsgrey
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Re: Ladder into Tower
« Reply #11 on: Apr 15th, 2003, 10:00am » |
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"[...]what is the longest ladder of length L[...]"
all ladders of length L are the same length...
Just a semantic quibble...
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