Card Staircase
If you take a pack of cards, you can slide them over to make
a kind staircase. What is the maximum horizontal distance the
top card can be from the bottom card?
The staircase will not have a constant slope. Use obvious
idealisations, this isn't a real engineering problem. You can
use as many cards as you want.
- Photino
a kind staircase. What is the maximum horizontal distance the
top card can be from the bottom card?
The staircase will not have a constant slope. Use obvious
idealisations, this isn't a real engineering problem. You can
use as many cards as you want.
- Photino
Labels: SharedPuzzle
posted by Rajesh Lal at
10:19 PM
5 Comments:
Would it be like:
N * X - (O(N))?
Where N is the number of cards, X is the length of the card, and O is the length of the card that overlaps the next?
An old one but a good one!
The top card on the staircase overlaps the next one by half a card. The second card (from the top) overlaps the third card by one third, the third one over the fourth by a quarter, and so on.
So, for a pack of 52 cards the bridge reaches a distance of:
(1/2 + 1/3 + . . . + 1/52) = 3.538...
With an infimite number of cards the bridge can reach infinity.
Glue the first card to the table, stagger and glue each other card on top of the previous. If the glue is good and the card stiffness is good you can go a lot futher that the 3.538(units?) Bobby suggested.
First anonymous - you are nearly right.
I guess you tried to remember the solution.
The nth card overlaps the one below it by L/(2n),
Where L is the length of a card.
For 52 cards the total overlap is:
(L/2)(1 +1/2 + 1/3 + 1/4 + 1/5 ....1/51) = 2.25941*L approx
Note that the 52nd card has no card to overlap.
As a previous poster said, if you have an infinite number of cards,
there is no limit to the amount of overlap.
To prove the equation, momemt or balance equations are applied
to the nth card.
If anyone asks, I'll provide a proof, but it will be awkard
because it really needs a diagram for it.
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