bonanova Posted February 18, 2008 Report Share Posted February 18, 2008 A vine has grown from the ground to a height of 15 feet on a tree in my back yard. I counted four times that the vine encircled the tree as it grew. I measured the circumference of the tree to be 2 feet. How long is the vine? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 18, 2008 Report Share Posted February 18, 2008 A vine has grown from the ground to a height of 15 feet on a tree in my back yard. I counted four times that the vine encircled the tree as it grew. I measured the circumference of the tree to be 2 feet. How long is the vine? 17ft Each time the encircles the tree it completes a helix. So there are 4 helices, with pitch of 15/4 ft around a circumference of 2 ft. If you could actually stretch one of these helices, then the length of the helix will be hypotenuse to the height and the circumference (Geometry 101). Thus Length of each Helix = ((3.75^2)+2^2)^0.5) = 4.25 Thus length of 4 such helices = 17ft Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted August 14, 2008 Report Share Posted August 14, 2008 L = (H^2 + C^2)^.5 L = the length of the vine for 1 revolution H = the height of 1 revolution (1/4 the total) C = the circumference So L = (3.75^2 + 6.2832^2)^.5 L = (14.0625 + 39.4786)^.5 L = 53.5411^.5 L = 7.3172 One reveolution is 1/4 of the way up, so. 4L = 29.2687 If the revolutions are evenly spaced I'd say the length of the vine is 29.27 feet long and maybe it's time to clean the yard. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 14, 2008 Report Share Posted August 14, 2008 17ft Each time the encircles the tree it completes a helix. So there are 4 helices, with pitch of 15/4 ft around a circumference of 2 ft. If you could actually stretch one of these helices, then the length of the helix will be hypotenuse to the height and the circumference (Geometry 101). Thus Length of each Helix = ((3.75^2)+2^2)^0.5) = 4.25 Thus length of 4 such helices = 17ft Agreed. I think there's a slightly easier way to compute it, which is:instead of dividing the length by 4, then multiplying the length of one helix by 4 at the end, I think it's easier to multiply the circumfrence by 4 in the beginning which gives you: sqrt(152+(2*4)2)=17 It's the same thing, just this way you don't have to deal with the fractions. Also, I had totally forgotten about that triplet. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 14, 2008 Report Share Posted August 14, 2008 120ft. mabe? Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted August 14, 2008 Report Share Posted August 14, 2008 Whoops... I thought the diameter was two feet I guess thats why you gotta read these things. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 14, 2008 Report Share Posted August 14, 2008 A vine has grown from the ground to a height of 15 feet on a tree in my back yard. I counted four times that the vine encircled the tree as it grew. I measured the circumference of the tree to be 2 feet. How long is the vine? I do not think the puzzle is solvable unless the diameter of the vine is known and then we make an assumption on the location of the neutral axis. This is because when you bend the vine around the tree some areas will be in tension and others in compression. The neutral axis, from where we calculate the true length will be neither in compression or tension. This would usually occur at the center of a round section when bent around a fairly large diameter. As the bending becomes more severe the neutral axis starts to shift towards the inside and can be as little as 1/3 the diameter. As an example if the tree were 24" in diameter and we wrapped a 1" round vine around it, the true length of the vine would be pi * (24 + 1), if we assume the neutral axis is at the center. This assumption on the neutral axis would be safe in this example and only rules of thumb tell us that when the ratios of the diameters D/d require an adjustment of the neutral axis position towards the inside. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 14, 2008 Report Share Posted August 14, 2008 I do not think the puzzle is solvable unless the diameter of the vine is known and then we make an assumption on the location of the neutral axis. This is because when you bend the vine around the tree some areas will be in tension and others in compression. The neutral axis, from where we calculate the true length will be neither in compression or tension. This would usually occur at the center of a round section when bent around a fairly large diameter. As the bending becomes more severe the neutral axis starts to shift towards the inside and can be as little as 1/3 the diameter. As an example if the tree were 24" in diameter and we wrapped a 1" round vine around it, the true length of the vine would be pi * (24 + 1), if we assume the neutral axis is at the center. This assumption on the neutral axis would be safe in this example and only rules of thumb tell us that when the ratios of the diameters D/d require an adjustment of the neutral axis position towards the inside. I think if we're going to neglect the irregularities of the tree (every tree I've seen tapers as it goes upwards) then we can also neglect the diameter of the vine. Especially with only one significant figure given for the circumference of the tree, I think the calculation is acceptably precise. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 14, 2008 Report Share Posted August 14, 2008 I think if we're going to neglect the irregularities of the tree (every tree I've seen tapers as it goes upwards) then we can also neglect the diameter of the vine. Especially with only one significant figure given for the circumference of the tree, I think the calculation is acceptably precise. I did the math so we can see the effect of the vine diameter: 1" dia, true length is 17.52 ft 3" dia, true length is 18.68 ft 6" dia, true length is 20.71 ft To me this is significant. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 14, 2008 Report Share Posted August 14, 2008 I did the math so we can see the effect of the vine diameter: 1" dia, true length is 17.52 ft 3" dia, true length is 18.68 ft 6" dia, true length is 20.71 ft To me this is significant. Good point. The relatively small diameter of the tree makes the diameter of the vine more important that I was thinking. Quote Link to comment Share on other sites More sharing options...
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bonanova
A vine has grown from the ground to a height of 15 feet on a tree in my back yard.
I counted four times that the vine encircled the tree as it grew.
I measured the circumference of the tree to be 2 feet.
How long is the vine?
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