The Cross Ratio - Numberphile
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Date: 2022-04-09
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Comments and reviews: 9
Michael
Geometric construction of cross-ratio, using compass and straightedge, is straightforward when one sees it is equivalent to the ratio of two rectangles' areas.
Rectangles can be constructably squared.
Translation and rotation are also c&s construction.
Ratio found by placing edge of each constructed square on a ray, with a corner concurrent and each external to the other, say.
A ray through similar corners, not on the base ray, is invariant to their ratio.
Do the same for any other line divided by a common projection point, and compare.
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Geometric construction of cross-ratio, using compass and straightedge, is straightforward when one sees it is equivalent to the ratio of two rectangles' areas.
Rectangles can be constructably squared.
Translation and rotation are also c&s construction.
Ratio found by placing edge of each constructed square on a ray, with a corner concurrent and each external to the other, say.
A ray through similar corners, not on the base ray, is invariant to their ratio.
Do the same for any other line divided by a common projection point, and compare.
reply
Wisnu
Cross ratio in geometry is beautiful. But using it in this context is like using an atomic bomb for killing an ant. The more efficient way is by counting the grass-line. The number of the grass-line between the ball and penalty box is the same with the number of the grass-line inside the penalty box. Since the penalty box is 16, 5m (18 yards, therefore the ball is about 33m (36 yards) from the goal. The articles you show in the beginning of thr videos is actually quite accurate. 30m is 33m rounded down. 40 yards is 36 yards rounded up.
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Cross ratio in geometry is beautiful. But using it in this context is like using an atomic bomb for killing an ant. The more efficient way is by counting the grass-line. The number of the grass-line between the ball and penalty box is the same with the number of the grass-line inside the penalty box. Since the penalty box is 16, 5m (18 yards, therefore the ball is about 33m (36 yards) from the goal. The articles you show in the beginning of thr videos is actually quite accurate. 30m is 33m rounded down. 40 yards is 36 yards rounded up.
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turtlellamacow
I'm missing some logical step here because I don't understand how the cross ratio theorem applies to this situation. He asserts that when four rays emanating from a point cut two other lines in a plane, the lengths of the segments into which the lines are cut have a fixed cross ratio, cool. How does one go from that to claiming that a line in real 3D space and its projection in a 2D photo obey the cross ratio rule (and the four cutting lines are parallel in real life, so they don't even emanate from a single point?
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I'm missing some logical step here because I don't understand how the cross ratio theorem applies to this situation. He asserts that when four rays emanating from a point cut two other lines in a plane, the lengths of the segments into which the lines are cut have a fixed cross ratio, cool. How does one go from that to claiming that a line in real 3D space and its projection in a 2D photo obey the cross ratio rule (and the four cutting lines are parallel in real life, so they don't even emanate from a single point?
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Koen
Great! I hate to spoil the party, but because I'd like to see how any measuring errors translate into an error estimate of the distance I did some calculations.
Assuming that the field lines are accurate enough not to contribute to the error, and assuming 1mm measuring error in each of the measurements, this leads to
a cross ratio of k=1. 251 +/- 5. 0%
Furthermore differentiation gave me
dx/dk = 999/(2. 7-2) so dx= 133 dk
(Using the calculated values)
Final result: x=33. 1 +/- 6. 7 m.
Hmmm.
reply
Great! I hate to spoil the party, but because I'd like to see how any measuring errors translate into an error estimate of the distance I did some calculations.
Assuming that the field lines are accurate enough not to contribute to the error, and assuming 1mm measuring error in each of the measurements, this leads to
a cross ratio of k=1. 251 +/- 5. 0%
Furthermore differentiation gave me
dx/dk = 999/(2. 7-2) so dx= 133 dk
(Using the calculated values)
Final result: x=33. 1 +/- 6. 7 m.
Hmmm.
reply
Josele
One intuition for the cross-ratio is to consider it as a ratio of ratios. If you modify slightly, this quantity is the ratio between to ratios with geometric interpretation. One is the ratio in which B is send to D as an homotecy of centre A. Whereas the other is the ratio in which B is send to D as an homotecy of centre C. And if the cross ratio is -1, it can be seen as if C is a centre of homotecy, then A is its inverse in the sense it has the same constant of homotecy but different sign.
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One intuition for the cross-ratio is to consider it as a ratio of ratios. If you modify slightly, this quantity is the ratio between to ratios with geometric interpretation. One is the ratio in which B is send to D as an homotecy of centre A. Whereas the other is the ratio in which B is send to D as an homotecy of centre C. And if the cross ratio is -1, it can be seen as if C is a centre of homotecy, then A is its inverse in the sense it has the same constant of homotecy but different sign.
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Mina
another easy way to calculate it is that the field ground is divided equally with those horizontal rectangles, the penalty box length fits for 6 rectangles (look at 14: 48 ) so the rectangle height is
16. 5 / 6 = 2. 75 m
then there is also 6 rectangles from the outside of the box to the ball meaning there is 12 rectangles from the goal line to the ball, the last step is just multiply 12 - 2. 75 = 33 m
it's not that accurate because the ball isn't precisely on the edge of a rectangle
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another easy way to calculate it is that the field ground is divided equally with those horizontal rectangles, the penalty box length fits for 6 rectangles (look at 14: 48 ) so the rectangle height is
16. 5 / 6 = 2. 75 m
then there is also 6 rectangles from the outside of the box to the ball meaning there is 12 rectangles from the goal line to the ball, the last step is just multiply 12 - 2. 75 = 33 m
it's not that accurate because the ball isn't precisely on the edge of a rectangle
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Jack
Nah bro. he did NOT have an amazing free kick ability, he used to take a lot and score rarely, but when he did it was spectacular. Its also NOT harder to create more spin with the outside of the foot, than the inside, its actually easier, however its harder to control (mainly because you generate more spin and the ball curves more, literally the opposite of what you said) you are wrong, I will happily show you on a pitch. Sorry but that bugged me.
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Nah bro. he did NOT have an amazing free kick ability, he used to take a lot and score rarely, but when he did it was spectacular. Its also NOT harder to create more spin with the outside of the foot, than the inside, its actually easier, however its harder to control (mainly because you generate more spin and the ball curves more, literally the opposite of what you said) you are wrong, I will happily show you on a pitch. Sorry but that bugged me.
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dimitris
Well you know the lenght of the whole field and you can measure the nuber of green stripes (darker and lighter ones) which are pretty much consistent in size and calculate the distance in terms of stipes. And you asume he was perpendicular to the goal line so assuming the stripes are perpendicular too you wouldn't deviate to much. This wouldn't as acurate but it would be accurate enough (approximately to +- 1. 8-2 meters)
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Well you know the lenght of the whole field and you can measure the nuber of green stripes (darker and lighter ones) which are pretty much consistent in size and calculate the distance in terms of stipes. And you asume he was perpendicular to the goal line so assuming the stripes are perpendicular too you wouldn't deviate to much. This wouldn't as acurate but it would be accurate enough (approximately to +- 1. 8-2 meters)
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Knut
Interesting find. I usually estimate this by thinking that the players are always about 9m from the ball to begin with, so it's easy to just add 16. 5+9 and make a reasonable guess for the distance between the 'wall' and the 16. 5m line. Or, if the shot is reasonably right on, I can do 11+2x9 and go from there.
It's just a rough guess, but I always thought this one was from like 32-33m based on that idea.
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Interesting find. I usually estimate this by thinking that the players are always about 9m from the ball to begin with, so it's easy to just add 16. 5+9 and make a reasonable guess for the distance between the 'wall' and the 16. 5m line. Or, if the shot is reasonably right on, I can do 11+2x9 and go from there.
It's just a rough guess, but I always thought this one was from like 32-33m based on that idea.
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