- 24 Feb '13 23:08After the try at cone volume, let me see if I can make any headway with another long standing question:

How high does one need to throw a basket ball so it passes through the hoop without touching it?

Looking at basketball rules, the hoop diameter is 18 (inches) and the mens' basketball has a 29.5 inch circumference, meaning its diameter is 29.5/pi (inches)

To make the contact equation more simple for me to think about, I am going to "thicken" the basketball hoop wire to a 29.5 inch circumference and assume the basketball is a point. This will not change the answer but now at least my "thick" thing is static.

The path of any basketball is a parabola, for my first try let me assume it is the very simple parabola y = K.x^2, where K is positive. This turns our scenario upside down because this parabola is a U. The cross section of the thickened wire is a circle nestling in the bottom of the parabola We are doing to define X=0,Y=0 to be the point at the bottom of the parabola, and we are going to see if there is a viable solution where the circle touches the parabola at just one point (0,0), if we can find such a solution with a K small enough for the parabola not to hit the back of the hoop then we have proved that there is a way to throw the basket ball at hoop_height+ball_radius + d and (only just) not touch the hoop, where d is an infinitesimal amount.

So the equation of the circular cross section of the "thick" hoop (that the parabola has to avoid) is

x^2 + (Y-R)^2 = R^2

and the equation of the parabola is Y=Kx^2

So the points at which the circle and the parabola touch are solutions to:

x^2 + (Kx^2 - R)^2 = R^2

multiplying out and rearranging:

X^2 + K^2.X^4 -2KX^2 + R^2 = R^2

X^2 + K^2.X^4 -2KX^2 = 0

X^2(1 + K^2.X^2 - 2K) = 0

this obviously has a solution at 0. Because of symmetry the parabola will intertect the circle either at one point (0,0) or at 3 points, and it can only intersect at 3 points if (1 + (KX)^2 - 2K) = 0 has 2 real solutions

we can rearrange that to (KX)^2 = (2K - 1)

and we see that this will only have 2 real solutions if K is greater than 0.5. As the parabola "widens" with K, the solution least likely to hit the back of the hoop is K=0.5, i.e. the parabola Y=0.5*X^2

To get a handle on whether this parabola misses the back of the hoop let us calculate the Y value of the parabola when X = 18-29.5/pi which is the first point that the parabola is at an X value occupied by the back of the hoop.

so Y = 0.5*(18-29.5/pi)^2

Y = about 37.06

since the thickened hoop is only 9.4 inches or so in diameter, we have shown that with this parabola, we are well "below" the back of the hoop by the time the ball gets to an X value where it overlaps the hoop X.

Therefore I have shown the ball needs to be thrown to hoop_height+ball_radius+delta, where delta is infinitesimally small. - 25 Feb '13 09:14

once again, there was a bit of a mistake there:*Originally posted by iamatiger***After the try at cone volume, let me see if I can make any headway with another long standing question:**

How high does one need to throw a basket ball so it passes through the hoop without touching it?

Looking at basketball rules, the hoop diameter is 18 (inches) and the mens' basketball has a 29.5 inch circumference, meaning its diameter is 29.5/pi ...[text shortened]... to be thrown to hoop_height+ball_radius+delta, where delta is infinitesimally small.

going from x^2 + (Kx^2 - R)^2 = R^2

this multiplies out to:

x^2 + K^2.X^4 -2KRx^2 + R^2 = R^2

(I forgot the R in -2KRx^2 last time)

x^2 + K^2.X^4 -2KRx^2 = 0

x^2(1 + (KX)^2 - 2KR) = 0

so we consider the number of real solutions of:

(1 + (KX)^2 - 2KR) = 0

which rearranges to

(KX)^2 = 2KR-1

clearly this only has 2 real solutions if 2KR is greater than 1

so for parabolas that don't hit the hoop we need:

2KR <= 1

i.e. K <= 1/(2R)

the equation for the circle at the back of the hoop is:

to be safe from the circle at the back of the hoop we want to be above the bottom of that circle (at Y >= 2R) when we reach the X value of the back (18-R)

and as larger Ks make the parabola sides steeper the parabola least likely to hit the back is where

K = 1/2R

recalling that the parabola is Y = KX^2

we need X^2/(2R) > 2R

(18-R)^2/(2R) > 2R

(18-R)^2 > (2R)^2

recalling that 2^ = 29.5/pi this evaluates to:

177 > 88

so our parabola is well clear of the back of the hoop - 27 Feb '13 22:35In fact that is more simply

(S-R)^2 > (2R)^2

(where S is the hoop diameter and R is the ball radius)

We square root this, with the additional knowledge that the ball radius is less than the hoop diameter, so:

S-R > 2R

S > 3R

Therefore we can guarantee a minimum height shot when this is true, in our case R is 29.5/(2*pi) = about 4.69 and S is 18, so we have the minimum parabola and can easily guarantee a minimum trajectory for all ball diameters up to 12 inches, and in practice a bit more as I have only done an approximate test for the parabola being clear of the hoop back (erring on the side of safety). - 03 Mar '13 18:02For the next stage of this puzzle, I am going to try to properly derive the equation for whether the parabola y = Kx^2 crosses a circle with the equation (x+a)^2 + (y+b)^2 = r^2

For the intersection points, the y value at a given x must be equal so we can form the equation:

(x+a)^2 + ( Kx^2+b)^2 = r^2

multiplying out

x^4 + (2Kb + 1)x^2 + 2ax + a^2-r^2 = 0

this quartic is a bit tricky to solve, let us assume it factorises into the product of 2 quadratics, so:

(x^2 + c^x + d)(x^2 + ex + f) = 0

multiplying that out:

x^4 + (c+e)x^3 + (d+f+ce)x^2 + (cf + de)x + df = 0

assuming the polynomial above is equal to our original one:

x^4 + (2Kb + 1)x^2 + 2ax + a^2-r^2 = 0

then the various factors of the powers of X must be equal, giving us 4 simultaneous equations which we rearrange to give the equations in squirly brackets:

c+e = 0 : {c = -e}

d+f+ce = 2Kb + 1 : {d+f-e^2 = 2Kb + 1 } : {d+f = 2Kb +e^2 + 1}

cf + de = 2a : {de - ef = 2a} : {d-f = 2a/e}

df = a^2-r^2

because we have definitions of d+f, and d-f we can use the identity:

(d+f)^2 - (d-f)^2 = 4df

and since we also know df = a^2-r^2, we can now form the equation

(2Kb +e^2 + 1)^2 - (2a/e)^2 = a^2-r^2

now defining E = e^2 we can turn this into:

(2Kb + 1)^2 + 2(2Kb + 1)E + E^2 - 4a^2/E = a^2-r^2

multiplying by E we get:

E^3 + 2(2Kb + 1)E^2 + ((2KB + 1)^2+r^2-a^2)E -4a^2 = 0

now we have reduced our quartic to a cubic - to be continued! - 14 Mar '13 22:29Aha - a bit of a brainwave!

we need the value of K for which the parabola:

y = K(x+c)^2

[i]touches[\i] a circle with the equation (x+a)^2 + (y+b)^2 = r^2

(I have made the parabola equation a bit more complex than before)

if we can find that, then we know for K values on one side of our value (keeping A constant) the parabola will fall short of the circle and not touch it.

Therefore we know two things:

a) there is a value X1 at which the Y values of the circle and the parabola are the same: this gives us the equation

(X1+a)^2 + (K(X1 + c)^2 + b)^2 = r^2

however we also know that:

b) the slopes of the parabola and the circle are equal at that point

it is known that the slope of the tangent line to a circle is given by:

dy/dx = (a-x)/(y-c)

the slope of our parabola:

y = K(x+c)^2

is got by expanding that and then differentiating

y = k(x^2 + 2xc + c^2)

dy/dx = 2k(x + c)

At the point where the two are equal:

2k(x + c) = (a-x)/(y-c)

subsituting x1 for x and K(x1 + c)^2 for y this gives:

2K(x1 + c) = (a-x)/( K(x1 + c)^2 + c)

which gives

2k^2(x1 + c)((x1 + c)^2 + c) = a-x1

or x1 = a - 2k^2(x1 + c)((x1 + c)^2 + c)

the equation we had before was:

(X1+a)^2 + (K(X1 + c)^2 + b)^2 = r^2

so we can substitute our value for x1, giving

(2a + 2k^2(x1 + c)((x1 + c)^2 + c)^2 + (K(X1 + c)^2 + b)^2 = r^2

now substituting Z = X1 + c we have:

(2a + 2k^2.Z)(Z^2 + c)^2 + (KZ^2 + b)^2 = r^2

hmm, not looking any easier, let me have a think