Dr. Dave's answers to frequently-asked questions (FAQs), mostly from the AZB discussion forum
for more information, see Sections 3.03 and 4.02 in The Illustrated Principles of Pool and Billiards
and Vol. I of the Video Encyclopedia of Pool Shots
What is a stun shot?
Stun means pure sliding, where there is no topspin or bottom spin. With a stun shot, the CB has no top or bottom spin when it hits the OB. For a straight in stun shot, the CB stops in place and transfers all of its energy to the OB. This is called a stop shot. With any other stun shot, with a cut angle, the CB heads in the tangent-line direction, regardless of the cut angle (see the 90-degree rule). A stun shot is sometimes called a "stop shot at an angle."
With a stun shot, the CB must be struck below center. The bottom spin will wear off (this is called drag) on the way to the OB. The farther the CB is from the OB, the lower and/or faster you need to strike the CB to prevent the CB from losing the bottom spin and starting to developing forward roll before hitting the OB. If the CB has any topspin when it hits the OB, the CB will follow forward. Here's a slow motion video showing how drag works with a stop shot:
Here's an example stun shot:
Here's a good drill for practicing stun shots from Vol. II of the Billiard University (BU) Instructional video series:
For more information, see "Billiard University (BU) - Part V: Stun Shots" (BD, December, 2013).
The following video illustrates stop, stun back, and stun forward shot variations and shows how you can use pages of a book to help ensure accurate tip contact heights on the CB:
from Patrick Johnson:
What is the 90-degree rule ?
The 90-degree rule states that for a stun shot, where the CB has no top or bottom spin at impact with the OB, the CB and OB separate at 90 degrees, regardless of the cut angle (except for a straight-in shot, in which case the CB stops in place). In other words, with a stun shot, the CB heads and persists along the tangent line direction, which is perpendicular to the line of centers (the line between the center of the ghost ball and OB).
The following diagrams illustrate the 90-degree separation angle between the CB and OB and how the tangent-line direction relates to other important CB position control reference lines.
Here's a convenient 1-page summary resource page summarizing all of the important points of the 90-degree rule.
Here are some relevant video demonstrations:
Here's a good drill for practicing stun shots: stop/follow/draw drill.
For more info, see "The 90° rule: Part I - the basics" (BD, January, 2004) and where the CB goes for different cases.
Concerning friction and english effects, see the following instructional articles:
"90° and 30° Rule Follow-up - Part II: speed effects" (BD, March, 2005)
"90° and 30° Rule Follow-up - Part III: inelasticity and friction effects" (BD, April, 2005).
"90° and 30° Rule Follow-up - Part IV: english effects" (BD, May, 2005).
"90° and 30° Rule Follow-up - Part V: the final chapter" (BD, June, 2005).
What is a stop shot, and why is it so effective?
A stop shot is a straight stun shot with no cut angle. With a stop shot, the CB stops in place after hitting the OB. To achieve this, the CB must have stun (no top or bottom spin) at OB impact. Sometimes a stun shot at an angle is also called a "stop shot" (or a "stop shot at an angle"), but this isn't strictly appropriate.
Stop shots can be very effective in run-out patterns since the CB position will be a known for the next shot.
Players can usually hit stop shots very reliably. One reason for this is that the stop shot is practiced and played often. Another reason is because of the physics of the shot. With a stop shot, the CB has a small amount of top/bottom spin over a fairly large distance, especially with faster-speed shots, and the CB stops within a very short distance (or no distance at all) with a small amount of spin (or no spin). For a more-technical proof, see:
TP B.8 - Draw shot physics
Specifically, see graphs "N" and "O" on page 17. These graphs show how much the CB draws back (d_draw) for draw/drag shots at different speeds and distances to the OB (d_drag). With the CB close to the OB (small d_drag), the CB draws back (d_draw>0). With the CB far from the OB (larger d_drag), and the same hit, the CB follows forward (d_draw<0). The graphs are very level near the stop-shot distance, especially for faster-speed shots (v3 > v1). That's the technical reason why stop shots are easier to control than precision draw or follow.
What is the math and physics explanation for why the CB must stop in place with a stop shot?
The Laws of Physics require that both energy and momentum be conserved during a collision. The physics and math details for a more general case (at any cut angle, including 0) in TP 3.1 - 90° rule.
For a straight shot (with no cut angle), the only way both energy and momentum conservation equations can be satisfied is if the CB stops and the OB moves away with the original CB speed (neglecting energy loss). It may seem hard to believe that the world needs to obey equations, but a better way to think about it is the equations just happen to described what actually happens in the real world.
Energy must be conserved because there is nothing adding or taking away energy from the balls (ignoring non-ideal collision losses). Momentum must be conserved because they are no external forces acting on the system (i.e., there is only an "internal force" acting between the balls during the collision). The only way both energy and momentum can be conserved is if all of the energy and momentum transfers from the CB and OB. If only a fraction of the momentum were transferred (with the CB retaining the other fraction), there would be a loss of energy, which is not possible. Here is the simple math showing why this is true:
m: ball mass (same for CB and OB)
v: original CB speed
vc: CB speed after collision
vb: OB speed after collision
Conservation of momentum: m*v = m*vc + m*vb
Conservation of energy: 1/2*m*v^2 = 1/2*m*vc^2 + 1/2*m*vb^2
These equations can be simplified by cancelling the equal massés, canceling the 1/2's, writing the squares (v^2) as multiplication (v*v), and assuming the initial speed is 1 (100% or full speed):
1 = vc + vb
1 = vc*vc + vb*vb
vc and vb must be numbers between 0 (no speed) and 1 (full speed). With a stop shot, the CB delivers all speed to the OB, so vc=0 and vb=1, and both equations are satisfied:
1 = 0 + 1 = 1
1 = 0*0 + 1*1 = 0 + 1 = 1
If the CB transferred only half of its speed to the OB, momentum is conserved, but energy is not:
1 = 1/2 + 1/2 = 1
1 = 1/2*1/2 + 1/2*1/2 = 1/4 + 1/4 = 1/2 (half of the energy is missing)!
The energy equation would fail for any fractions you try to use for vc and vb because when you square fractions, you get smaller fractions, which can't add to 1.
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