Here we have a point-like ball moving inside a (soft, elastic) circular boundary i.e. a circular billiard table. Interestingly, Charles L. Dodgson, (Lewis Carroll of Alice's Adventures in Wonderland fame), once published a set of rules for a two-player game of circular billiards [1]. However, this is predated somewhat by Alhazen / Ptolemy [2].
This post follows the same approach as the previous post, but with the V-wedge replaced with a full circle of radius R:
q is the distance from a point (x,y) to the circular surface, measured along the normal to the surface.
Hence if a point-like ball is at a point (x,y) outside the circular perimeter, there is an elastic restoring force (acting towards the centre of the circle) given by kq, where k is the spring constant of the wall of the circle. It is easy enough to work out q, and thus the accelerations along x and y:
where m is the mass of the ball.
Acceleration due to gravity is directed purely along the negative y-axis. We assume that the radius of the circle is small compared to the radius of the earth (i.e. g is constant).
The computer's squarers yield the square of the input voltage divided by ten - both quantities Q and q end up divided by the square root of ten; this comes out in the wash of the divider, however. The only point to remember is that the voltage corresponding the the radius R must also be divided by √10 also (before it gets subtracted from Q/√10 to yield q/√10). Here's the patching:
No gravity case
We simply set g = 0.
Now the ball doesn't move (there is no initial force). However if we set the initial velocity to something non-zero the point-like ball moves across to the circular wall, and rebounds. This is most interesting perhaps if we start the ball from someplace other than the origin - otherwise the ball simply oscillates back and forth through the origin. The following plots show the trajectory for the ball starting at (1,1) with an initial velocity of 3 m/s along the positive x-axis. -R√10 is set to -0.632 V, corresponding to a billard table radius R of 2 (metres).
In all cases, the trajectory was plotted for eight minutes (integrators have 1 second time constants). Different coloured plots are for different wall spring constants: (a) k = 0.2 N/m (black line), (b) k = 1 N/m (red line), (c) k = 10 N/m (these spring constants assume a 1 kg mass).
The results indicate that is indeed not possible for the trajectory of a moving ball within a circular billiard table to cover the whole table completely. Unless of course...
Gravity Case
...we tilt the table. In other words we have a non-zero gravitational force (here along the negative y-axis).
We start with the ball at (1,1) as before, but with zero initial velocity. Acceleration due to gravity (g) is 10 m/s2. The following result is for k = 10 N/m (eight minutes of trajectory).
A lesser tilt also gives extensive coverage of the (lower half) of the table...here g = 2 m/s2 (about an 11° slope). Everything else as before:
An even lesser tilt (g = 0.5 m/s2 - about a 2.8° slope on planet Earth) still yields good coverage - albeit at a slower pace:
Finally, g = 0.1 m/s2 (about a 0.57° slope on planet Earth - and surely hardly noticeable in a darkened pool hall) yields the result below. This result is suspect to some degree however, since the ball passes higher than its starting height - suggests some issue with the computation - e.g. an offset null somewhere (e.g. squarer outputs), in effect inputting energy into the system:
References
[1] Dodgson, Charles Lutwidge ('Lewis Carroll'). Circular Billiards for two players. 1890.
[2] Alhazen's problem corresponds to finding the point on the edge of a circular billiard table at which a cue ball at a given point must be aimed in order to bounce off the edge of the table and strike another ball at a second given point, named after the Arab scholar Abū ʿAlī al-Ḥasan ibn al-Ḥasan ibn al-Haytham (c. 965 AD – c. 1040 AD), although it was first formulated by Ptolemy even earlier, in 150 AD.
This post follows the same approach as the previous post, but with the V-wedge replaced with a full circle of radius R:
q is the distance from a point (x,y) to the circular surface, measured along the normal to the surface.
Hence if a point-like ball is at a point (x,y) outside the circular perimeter, there is an elastic restoring force (acting towards the centre of the circle) given by kq, where k is the spring constant of the wall of the circle. It is easy enough to work out q, and thus the accelerations along x and y:
where m is the mass of the ball.
Acceleration due to gravity is directed purely along the negative y-axis. We assume that the radius of the circle is small compared to the radius of the earth (i.e. g is constant).
The computer's squarers yield the square of the input voltage divided by ten - both quantities Q and q end up divided by the square root of ten; this comes out in the wash of the divider, however. The only point to remember is that the voltage corresponding the the radius R must also be divided by √10 also (before it gets subtracted from Q/√10 to yield q/√10). Here's the patching:
No gravity case
We simply set g = 0.
Now the ball doesn't move (there is no initial force). However if we set the initial velocity to something non-zero the point-like ball moves across to the circular wall, and rebounds. This is most interesting perhaps if we start the ball from someplace other than the origin - otherwise the ball simply oscillates back and forth through the origin. The following plots show the trajectory for the ball starting at (1,1) with an initial velocity of 3 m/s along the positive x-axis. -R√10 is set to -0.632 V, corresponding to a billard table radius R of 2 (metres).
In all cases, the trajectory was plotted for eight minutes (integrators have 1 second time constants). Different coloured plots are for different wall spring constants: (a) k = 0.2 N/m (black line), (b) k = 1 N/m (red line), (c) k = 10 N/m (these spring constants assume a 1 kg mass).
The results indicate that is indeed not possible for the trajectory of a moving ball within a circular billiard table to cover the whole table completely. Unless of course...
Gravity Case
...we tilt the table. In other words we have a non-zero gravitational force (here along the negative y-axis).
We start with the ball at (1,1) as before, but with zero initial velocity. Acceleration due to gravity (g) is 10 m/s2. The following result is for k = 10 N/m (eight minutes of trajectory).
A lesser tilt also gives extensive coverage of the (lower half) of the table...here g = 2 m/s2 (about an 11° slope). Everything else as before:
An even lesser tilt (g = 0.5 m/s2 - about a 2.8° slope on planet Earth) still yields good coverage - albeit at a slower pace:
Finally, g = 0.1 m/s2 (about a 0.57° slope on planet Earth - and surely hardly noticeable in a darkened pool hall) yields the result below. This result is suspect to some degree however, since the ball passes higher than its starting height - suggests some issue with the computation - e.g. an offset null somewhere (e.g. squarer outputs), in effect inputting energy into the system:
References
[1] Dodgson, Charles Lutwidge ('Lewis Carroll'). Circular Billiards for two players. 1890.
[2] Alhazen's problem corresponds to finding the point on the edge of a circular billiard table at which a cue ball at a given point must be aimed in order to bounce off the edge of the table and strike another ball at a second given point, named after the Arab scholar Abū ʿAlī al-Ḥasan ibn al-Ḥasan ibn al-Haytham (c. 965 AD – c. 1040 AD), although it was first formulated by Ptolemy even earlier, in 150 AD.
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