PAP Identification Method (& Terminology)

[Mike White]

How long ago did this method of identifying a bowlers pap location get started, and is there time to change the method?

The method of over X and up Y the way it's implemented to me is like sticking with the english method when clearly the metric system is better.

The X and Y values simply don't translate into anything useful other than identifying a location.

In my opinion the PAP location should be the equivalent to the North Pole, and the line from the PAP to the center of the fingers (not including the thumb) should be the equivalent to the Prime Meridian.

Measure from the PAP to the center of the fingers gives you a longitude, latitude value, longitude being 0 degrees.

The equator is always 6.75" from the PAP or North Pole.

Positioning the PSA, and PIN make sense when referenced to the Prime Meridian.

When you release the ball, is the PIN ahead of your fingers, or behind? The longitude coordinate of the PIN would answer that question. It's similar to "Does the ball have finger or thumb weight?" in the "old" days.

As an example, if your fingers are 3.5" latitude, and you have 40 degrees of axis rotation, your fingers are almost directly behind the ball at release. Off by just 6 2/3 degrees.

[bowl1820]

Update: The links to mikes images appear to no longer work. So some images are missing.

How long ago did this method of identifying a bowlers pap location get started, and is there time to change the method?

When did they start using it? Probably a little while after they discovered that they needed to know where the PAP was.
Is there time to change? Nope that ship sailed already.

The method of over X and up Y the way it's implemented to me is like sticking with the english method when clearly the metric system is better.

The X and Y values simply don't translate into anything useful other than identifying a location.

That is all that's needed, it locates the PAP.

In my opinion the PAP location should be the equivalent to the North Pole, and the line from the PAP to the center of the fingers (not including the thumb) should be the equivalent to the Prime Meridian.

Measure from the PAP to the center of the fingers gives you a longitude, latitude value, longitude being 0 degrees.

The equator is always 6.75" from the PAP or North Pole.

Positioning the PSA, and PIN make sense when referenced to the Prime Meridian.

When you release the ball, is the PSA ahead of your fingers, or behind? The longitude coordinate of the PSA would answer that question. It's similar to "Does the ball have finger or thumb weight?" in the "old" days.

As an example, if your fingers are 3.5" latitude, and you have 40 degrees of axis rotation, your fingers are almost directly behind the ball at release. Off by just 6 2/3 degrees.

This is just a confusing mess.

But if you want a real good opinion go over to the forum on http://www.bowlingchat.net/ and post your idea there.
Mo Pinel one of the top ball and drilling experts is on there all the time. Get his opinion on your idea.

[billf]

Not that we are trying to send people away but Mo will know better than the rest of us if it's feasible and have no issues explaining why if it's not.
Who knows, maybe Mike is onto the next new thing in drilling.

[bowl1820]

Measure from the PAP to the center of the fingers gives you a longitude, latitude value, longitude being 0 degrees.

Also-
What is the center of the fingers?
Your still going to need to use over X and up Y coordinates to locate the pap.
What about when you project back from the pap to your grip center line to layout the finger and thumb holes?
Where ever the "center" of the fingers is, it will then will require more measurements to locate the finger holes than is needed now.

[Mike White]

I am in the process of brushing up on my Linear Algebra to identify locations on a sphere after being turned on the X, Y, and Z axis.

Just to be sure my assumptions are correct. Is distance from the PIN to the PSA is always 6 3/4" ?

[bowl1820]

I am in the process of brushing up on my Linear Algebra to identify locations on a sphere after being turned on the X, Y, and Z axis.

Just to be sure my assumptions are correct. Is distance from the PIN to the PSA is always 6 3/4" ?

Yes 6 3/4", 1/4 of the balls circumference.

If you have a ball with a unmarked MB you draw a line from the pin through the CG, then measure from the pin 6 3/4" and mark that spot on the line, thats the mass bias (PSA)

A little extra info for you.
The neutral position is (4 1/8" from the x,y,and z axis)

This is using a Determinator.
Measuring the X, Y, and Z Axes of a Bowling Ball.
1. Place the ball on the deTerminator with the pin aligned with the eyehole of the support arm.
2. Turn the deTerminator on and trace the ball path through the eyehole. The ball will eventually spin about a fixed axis (this axis runs through the bowling ball – from the surface, the ball will spin about a point) – this is known as the Preferred Spin Axis or PSA. This PSA represents the Y or High RG Axis of the bowling ball. Mark the spot through the eye hole (dot) and the top of the arm, creating a small circle about the PSA.
3. Using the pro-sect, draw a line from the PSA towards and through the manufacturer pin of the bowling ball. Mark the point 6 3⁄4” from the PSA along the line – this will represent your X axis.
4. Now draw a Perpendicular Line from the X axis toward the ‘equator’ of the ball. Mark a spot 6 3⁄4” down the line from the X axis - this will give you the Z or intermediate RG axis.
5. To ensure the accuracy of this location, take your pro-sect and measure perpendicular from the X axis to Y axis line toward the Z axis. This line should directly intersect where your Z axis is marked and be 6 3⁄4” in length. If they do intersect and your line is 6 3⁄4”, you have correctly found all three axes. If not, be sure to check that your lines are 6 3⁄4” long (this is 1⁄4 of a bowling ball’s circumference) and perpendicularly intersect at the axis locations.

http://www.bowl.com/uploadedFiles/Eq...onAnalysis.pdf

[Mike White]

Can you clarify what you mean by neutral position?

[bowl1820]

Can you clarify what you mean by neutral position?

The Neutral Position is the point that is equidistant from all three major axes.

It has to do with the determinator and measuring the spin time of a ball.

spin time is a real world measurement of the complex relationship between RG, Diff, Int Diff, Diff Ratio. It quantifies the strength of the PSA, which affects the quickness of the reaction to friction on the lanes.

The Neutral Position

Here's a little mathematical proof for you.

The image below shows us a triangle drawn on a sphere:
Image

The law of sines states (as found on wikipedia):

Sin a / Sin α = Sin b / Sin β = Sin c / Sin γ

Where γ is the angle created at vertex C, and a, b, c are the angles created by the arcs, relative to the center of the circle. Since the arc c = 6.75", we know the angle at the center of the circle is 90°.

If α = β = 45° we can find γ, using the second spherical law of cosines (also found wikipedia).

Cos γ = - Cos α * Cos β + Sin α * Sin β * Cos c
Cos γ = - Cos 45° * Cos 45° + Sin 45° * Sin 45° * Cos 90°
Cos γ = - Cos 45° * Cos 45° (since Cos 90° = 0)
Cos γ = -0.5
∴ γ = ArcCos (-0.5) = 120° [ArcCos is the Inverse Cos operation]

Using the law of sine (see above), we can find solve for a, as follows:

Sin a = Sin c * Sin α / Sin γ

Since c = 90°, Sin c = 1, so this formula becomes:

Sin a = Sin α / Sin γ
Sin a = Sin 45° / Sin 120°
a = ArcSin (Sin 45° / Sin 120°) [ArcSin is the Inverse Sin operation]
∴ a ≈ 55° (54.7356° to 4 decimal places)

All that's left is to find the length of the arc a, we first convert a to radians, a

a = a * ╥ / 180 = (55 * ╥ / 180)

Then we multiply a by the radius of the ball to get the arc length, which is 13.5 /╥

∴ a = (55 * ╥ / 180) * (13.5 / ╥) = 4.125"

If you use the EXACT angle for a, it comes out to 4.105", which for our purposes, is the same is 4.125", since the difference is a little over 1/64".

[billf]

^^^now that makes more sense to me

[Mike White]

The Neutral Position is the point that is equidistant from all three major axes.

That was all the info I needed. I wasn't sure if neutral meant something about the balance (static or dynamic)

Finding the distance along the sphere from an axis to the neutral position can be done with some simple algebra, and just a hint of trigonometry.

Imagine our sphere has a radius of 1. (it makes the math much easier and we'll scale it back up at the end).

The coordinates for the three main locations in your picture could be:

A (1,0,0)
B (0,1,0)
C (0,0,1)

Three points make a plane, and those 3 points make an equilateral triangle on that plane. To find the center of that triangle we average the coordinates. Lets call that point D.
D = (A + B + C) / 3 = (1/3, 1/3, 1/3)

The important thing we get from that is the x, y, and z coordinates of point D all have the same value.

D however is not on the surface of the sphere. We would need to project a line from the center of the sphere thru point D and find where it touches on the sphere. Lets call that point N. N = (Nx, Ny, Nz)

We do know that the distance from the center to N is 1 since we selected a radius of 1. Using the distance formula:

sqrt(Nx^2+Ny^2+Nz^2) = 1

square both sides

Nx^2 + Ny^2 + Nz^2 = 1

Remember that Nx = Ny, and Ny = Nz

Nx^2 + Nx^2 + Nx^2 = 1

3(Nx^2) = 1

Nx^2 = 1/3

Nx = sqrt(1/3)
Ny = sqrt(1/3)
Nz = sqrt(1/3)

The formula to convert from spherical coordinates to (x, y, z) type coordinates is.

X = cos(theta)*sin(phi)
Y = sin(theta)*sin(phi)
Z = cos(phi)

Where theta is similar to a longitude angle, and phi is similar to a latitude angle, except phi = 0 at the north pole and 90 degrees at the equator.

since Nz = sort(1/3) = cos(phi) we need to know what angle phi would give us a value sqrt(1/3) from the cosine function. 54.736 degrees gives that value.

Now we scale up to the bowling ball size with a simple proportion problem.

54.736 / 90 = v / 6.75

v = 4.105