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| daewoo_lowrider Quote | Reply | | Please help - Rugby Goal Kicking Problem posted on: 5/11/2007 10:56:04 PM Hi, can someone please help me with this problem? It states: Suppose a try is scored at T, then the kicker can kick to the posts from anywhere along the line TF perpendicular to the try line PQ. (Refer to Figure 1) Where is the best place to kick from? Many kickers seem to think that the further they go back (within their kicking limit) the better. But this is not necessarily the case. The best position for the kick to be taken is clearly where angle ACB subtended by the posts along TF is a maximum. Here is the diagram:  Basically, I think I have to use trig, and a circle subtending the goal posts, but I'm stuck at that, I have hardly any measurements and angles (the distance between the goal posts is 6.5m. I'm guessing I have to make an assumption as to how far away the try was scored from the posts and go from there. Thanks for any help, Daniel |
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cramwit
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Please help - Rugby Goal Kicking Problem
replied on: 5/12/2007 3:32:46 AM Look at the comparison of angles: ATB = 0 ACB = mid sized (of these 3) AFB = largest angle (of these 3) The distance to the goal line is shortest from ATB & sequentially longer up to AFB so if distance is a concern then that would work conversely. Obviously you probably don't want BC or B? to exceed your kicking distance. As far as you, the kicker are concerned the distance between goal posts change depending on your position along TF. To calculate the particular 'window' angle you subtract the ACT angle from the BCT angle, or putting it as a variable; subtract A?T angle from the B?T angle. The AT, BT & AB segment distances never change. But the CT or rather ?T segment does change. arctan(AT/T?) = angle A?T arctan(BT/T?) = angle B?T arctan(BT/T?) - arctan(AT/T?) = angle A?B You are looking for the maximum/widest angle of A?B; that is the widest window for you as the kicker. It is probably maximized where segment AB = AC or AB = A? not sure how to determine that for sure though. as C->T ACB or A?B goes to zero as C->F ACB or A?B goes to zero If you think of A & B fixed & C floating around freely in space optimally it is at the mid point of segment AB making A-C-B a flat or 180 degree angle. essentially the goal already being made before kicking. But you are restricted to remain on TF. I think it is where AC also equals 6.5 meters but since i can't explain why it is only a guess. Maybe that maximizes the area to perimeter ratio, but all the kicker cares about is maximizing the particular kicking angle. You could find it with a compass by putting the point at A & drawing a circle from B until it crosses TF which would be C I will have to think about a/the reason why that would be correct, for now it is just a guess. |
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cramwit
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Please help - Rugby Goal Kicking Problem
replied on: 5/15/2007 4:55:30 AM Just throwing out some novice thoughts, I was thinking there must be some relationship between a triangle sidelength & its perimeter & its angle to the total 180 degrees. It wasn't as straightforward as i hoped/thought. If you take a given sidelength & a given perimeter length you essentially have the mechanics used to draw an ellipse. ie. two pins & a given length loop of string. So i guess you could say the sidelength to perimeter length has some kind of elliptical relationship to its corresponding angle based on the other side lengths. In euclidean geometry any single triangle's side length can not exceed 1/2 the total perimeter. If it exactly equals 1/2 of the perimeter it will always be 180 degrees (pi/2), & the other two angles are zero, & the corresponding/opposite angle is maximum. Also [euclidean] if the side length is essentially zero of the total perimeter then the other sides overlap & draw a circle. And the corresponding angle of the zero side is essentially zero, or minimum. Between these two extremes the maximum corresponding/opposite angle to side 'n' is where the other two side are equal, creating an isosceles triangle. In this case you have two competing functions, minimal perimeter length & equivalence of the other two sides i think you are looking for the minimum perimeter that gives the closest to equal other sides. The intersection between these functions given the stipulation of the 3rd point is on FT is what should give you the maximum opposition angle to segment AB. The two other sides can be algebraically/trigonometrically derived plugging in variable side lengths, then summing them to get perimeter. It might be that if you create both functions algebraically the variable elements cancel & you come up with seg length AB as the answer for maximum opposition angle, but honestly i am really grasping/guessing at all of this. |
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cramwit
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Please help - Rugby Goal Kicking Problem
replied on: 5/15/2007 5:26:35 AM perimeter = AB + AC + BC AB = 6.5m AC = sqrt( (AC)^2 + (CT)^2 ) BC = sqrt( (BT)^2 + (CT)^2 ) sort of lim AC -> BC & lim ( AB + AC + BC ) -> 2AB or lim ( AC + BC ) -> AB lim AC -> BC = lim sqrt((AC)^2+(CT)^2) -> sqrt((BT)^2+(CT)^2) [don't know if any of that can be canceled in a limit] lim AC + BC -> AB = lim sqrt((AC)^2+(CT)^2) + sqrt((BT)^2+(CT)^2) -> AB probably making a hash out of this. maybe it works better using trig angle functions for a minimum angle 'C' |
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