Acceptance Angle of a Hoop
To save others working through the geometry and to provide thinking material, I give a geometric analysis of angles from which you can run a hoop (the acceptance angle) given many simplifications and assumptions.
Balls are treated as hard, non-rolling, perfectly elastic spheres. The hoop uprights are hard, perfectly rigid and elastic. Consequently what is shown is only a simple geometric interpretation.
Running a Hoop without Collision
The normal view of the maximum acceptance angle, θ, which a hoop can be run from is shown opposite.
It is where the ball just misses the near and far uprights. It is a function of the upright and ball diameters and the uprights' separation. The magnified diagram below gives the trigonometry. The acceptance angle is theta, θ.
Here Rb is the radius of the ball, typically 46.0375 mm.
Ru is the radius of the upright, nowadays this can be 5/8" or ¾". Taking 5/8" Ru = 7.9375 mm
The right-angle magenta triangle gives the angle θ in terms of the distances A and B.
From inspecting the diagram B = Rb + Ru, hence B = 53.975 mm. A is the distance from the centre of the upright to the centre of the hoop which is A = Rb + Ru + Gap/2; this is also A = B + Gap/2. Gap is the total clearance between the ball and the hoop uprights (not labelled in the diagram). For a hoop set to 1/8" (3.175 mm) we get A = 53.975 + 3.175/2 = 55.5625.
Trigonometry gives cos(θ) = B / A = 53.975 / 55.5625 = 0.97142857
Taking the inverse cosine gives:
θ = 13.73º
or as a single formula:
θ = arccos( (Rb + Ru) / (Rb + Ru + Gap/2) )
What happens when the ball makes collisions with one or more uprights?
Running a Hoop with a Single Collision
Firstly to introduce the method of constructing the diagrams:
In A we are considering a particle travelling along the green line and hitting the blue surface and bouncing off it. Given our assumptions, the angle it comes in on is the same as the angle on which it leaves. If we drop the surface down to where the black line is and use a ball we get the situation in B. The ball meets the surface and bounces off it, again where the incoming angle equals the outgoing angle. We can still consider the green track with respect to the centre of the ball and a displaced blue boundary equivalent to the radius of the ball.
Given that the upright is circular we can draw the blue circle (radius = ball radius plus upright radius) which shows the closest that the centre of the ball can approach the upright (C).
For a ball which strikes the upright at an arbitrary angle (D), a radius is extended to where the centre of the ball hits the blue circle and the incoming angle is mirrored through this radius to give the outgoing path. This form of construction has been used in the following diagrams.
In the diagram above a ball just clears the right-hand upright and strikes the left-hand upright; the full Gap of the hoop is on the right of the ball. As before below is a magnified image.
The difference here is that the ball is not in the centre of the hoop but touching the left-hand upright. Hence the full Gap contributes to the hypotenuse and consequently the formula for θ becomes:
θ = arccos( (Rb + Ru) / (Rb + Ru + Gap) ).
Does it make much difference?
cos(θ) = B / C = 53.975 / (53.975 + 3.175) = 0.9444
arccos(θ) = 19.19º
... around 5.46º (19.19 - 13.73) is worth having in this artificial model! The take-home message here is not to aim at the centre of the hoop but so the ball just misses the nearest upright - but we knew that.
If one collision increases the acceptance angle what about multiple collisions?
Some multiple collisions are unwelcome (opposite) but what about those where a ball goes through a hoop?
Here I show a non-symmetrical case with two collisions. Assuming the ball approaches from the top, it kisses the left-hand upright, rebounds on to the right-hand one and exits.
Below is a magnified picture:
The top ball arriving perpendicular to the hoop is offset from the centre of the hoop - this is aiming straight toward the hoop but not at the centre of the hoop. It cannot move much to the left without the ball bouncing back into the hoop from the right upright Because of the curvature of the blue 'rebound' circles, a small change in position produces a big deviation. It does show that if you aim perpendicular to the plane of the hoop but are slightly more than half the Gap left or right (i.e. there is an offset) you will get through with a double rebound.
Most things being reversible, reading the diagram from the bottom to the top; if you shoot the bottom ball at an angle into the hoop aiming away from the centre of the jaws you will get the ball emerging at a range of angles.
If there is the single collision as in the section above the ball will leave at the maximum angle but in the opposite direction. As in the diagram immediately above, there is an angle where a double collision will 'straighten' the emerging ball.
Out of curiosity are higher numbers of collisions possible? Yes, but the conditions become increasingly narrow. Three collisions are shown below:
Simple formulae have been presented to gauge the acceptance angles of hoops under artificial assumptions. The acceptance angle is significantly increased when one takes into account a single collision on an upright. For a hoop with 5/8" uprights and 1/8" gap the acceptance angle without touching an upright is 13.73º whereas that acceptance angle is 19.19º if the ball is first bounced off the far upright. (The angle is measured from a marginally different origin in the second instance).
Because of the geometry, double and higher collisions with uprights only occur if the ball is well within the acceptance angles above and do not increase them.
In the real world the hoop and ball are not perfectly elastic and hard hence the ball will not deflect at the same angle as it collided and it will lose energy. For a championship ball the specified rebound height of a ball dropped from 60" is between 31-37", so 38-48% of the energy is lost! Hoops are not totally rigid hence will deflect when bashed by a ball altering the geometry. Finally the key to running a hoop if there is a collision is imbuing the ball with top-spin. On collision linear momentum is lost but not angular; hence after a collision the spin of a ball may drag the ball through a hoop.
All rights reserved © 2016-2017