605
   

NeoPets Riddles (Lenny Conundrums) and Answers Here

 
 
K311Y
 
  1  
Reply Wed 24 Jun, 2009 05:56 pm
@jjbuttcrack,
Ok, so maybe this is the formula...

143.75-40.82=102.93
102.93 x 15 = 1543.95

143.75 is 2300/16 and is the amount of string used per revolution
40.82 is circumference

So Im thinking we should deduct the 40.82 cm per revolution since that is a given. 102.93 is (to my brain anyway) the extra string used per revolution as it is stretched down the pole.
Leopard
 
  1  
Reply Wed 24 Jun, 2009 05:57 pm
@Leopard,
if we did tightly wrap it around the pole the string would go around 56.3 times. :S
0 Replies
 
jjbuttcrack
 
  1  
Reply Wed 24 Jun, 2009 05:58 pm
guys guys guys -- look at my first post (#3 on 606) -- i show that it canNOT be tightly wrapped. and that means that the thickness of the rope is not necessary to solve it. the thickness would not matter at all. really, its the tightness of the spiral that will define the length of the pipe the rope will cover.
0 Replies
 
warhammer11
 
  1  
Reply Wed 24 Jun, 2009 05:58 pm
can't we try it out manually
0 Replies
 
starleogirl09
 
  1  
Reply Wed 24 Jun, 2009 05:59 pm
@K311Y,
Why multiply by 15?
Leopard
 
  1  
Reply Wed 24 Jun, 2009 05:59 pm
So the pole is 102.93 cm long per revolution?
0 Replies
 
jjbuttcrack
 
  1  
Reply Wed 24 Jun, 2009 06:00 pm
i was thinking that too kelly -- but it somehow seemed wrong to me... but if you got there as well, maybe there's something to it.
Leopard
 
  1  
Reply Wed 24 Jun, 2009 06:01 pm
@jjbuttcrack,
ok, i submitted it.
0 Replies
 
K311Y
 
  1  
Reply Wed 24 Jun, 2009 06:03 pm
@starleogirl09,
I multiplied by 15 because that is how many spaces are between each coil. I drew a diagram. The string starts at the very top and ends at the very bottom, so you dont really have any extra string for those revolutions.
starleogirl09
 
  1  
Reply Wed 24 Jun, 2009 06:04 pm
Are we taking into account that each time the string goes around the pole, it would be more than the actual circumference because it is loosely wrapped?
0 Replies
 
wertyiu102
 
  1  
Reply Wed 24 Jun, 2009 06:05 pm
i got either 11 or 177 but, no decimals, round with the closest
0 Replies
 
starleogirl09
 
  1  
Reply Wed 24 Jun, 2009 06:05 pm
@K311Y,
I get it--thanks!
0 Replies
 
edstock
 
  1  
Reply Wed 24 Jun, 2009 06:05 pm
@warhammer11,
warhammer11 wrote:
i believe the string is coiled around the cylinder kinda of like the problem on this page :http://mathforum.org/library/drmath/view/55156.html


I think the approach of rolling it out into a triangle is exactly right. We are given L, the length of the string, D, the diameter of the cylinder, and N, the number of turns, and asked to solve for P*N.

Use the Pythagorean theorem for the relationship between the sides of a right triangle and out pops the answer.
wertyiu102
 
  1  
Reply Wed 24 Jun, 2009 06:06 pm
@edstock,
umm.......... DOes p= pi or pythagoream
Firestar624
 
  1  
Reply Wed 24 Jun, 2009 06:07 pm
@edstock,
Yep, I used that and just got my answer. But mines RLY big. Is urs?
K311Y
 
  1  
Reply Wed 24 Jun, 2009 06:08 pm
@edstock,
See, I thought that too, but there is no pitch to plug into that formula.
edstock
 
  1  
Reply Wed 24 Jun, 2009 06:08 pm
@wertyiu102,
wertyiu102 wrote:
umm.......... DOes p= pi or pythagoream


P = pitch, the height of each revolution of the string around the coil or cylinder. P*N is just the entire height of the cylinder or coil.
0 Replies
 
wertyiu102
 
  1  
Reply Wed 24 Jun, 2009 06:09 pm
i don't like those holdon! you have to wait 87654321 seoncd before you post again messages
0 Replies
 
edstock
 
  1  
Reply Wed 24 Jun, 2009 06:10 pm
@K311Y,
Quote:
See, I thought that too, but there is no pitch to plug into that formula.


We don't need to know the pitch, we need to solve for the product P*N, but not by multiplying P and N. We use the lengths of the other two sides and calculate the length of the vertical side using the Pythagorean Theorem.
edstock
 
  1  
Reply Wed 24 Jun, 2009 06:11 pm
@Firestar624,
Quote:
Yep, I used that and just got my answer. But mines RLY big. Is urs?


Very big - the wrappings don't use much of the length of the string, as the height is so much bigger than the diameter.
0 Replies
 
 

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