Rules for speed of CB after hit?

Billy_Bob

Billy_Bob

AzB Silver Member
Silver Member
Is there any kind of "rule" or rule of thumb for how much energy will be used up when making a hit? (How much energy CB will lose.)

Like the hit itself;
Full ball hit
3/4 ball hit
1/2 ball hit
1/4 ball hit
thin hit

And then rails;
hits one rail and loses x amount of energy (After contacting object ball).
hits two rails and loses x amount of energy.
hits three rails and loses x amount of energy.

So say I had a half ball hit and the cue ball was going to hit 3 rails before traveling to where I wanted it to go. Half ball hit would remove x amount of speed, 1st rail x, 2nd x, 3rd x, etc.
 
Full ball hit - 100%
3/4 ball hit - 75%
1/2 ball hit - 50%
1/4 ball hit - 25%
thin hit - 0-10%

one rail - about 40% depends on the rail
 
basic physics (i'll leave out friction which is nearly negligible):

if you take the angle of the hit (half ball is 45', full-on 0', you get the point), the cosine of that angle represents how much force will transfer to the object ball. the cosine of 0' is 1 (or 100%), and the cosine of 90' is 0. now, a half-ball hit (45') results in about 70.7% transfer, leaving 29.3% (minus a little friction) with the cueball. so it's not a linear thing, mnShooter's post is wrong.

now, on rails. they're all a little different. your guess is as good as mine :)

-s
 
steev said:
basic physics (i'll leave out friction which is nearly negligible):

if you take the angle of the hit (half ball is 45', full-on 0', you get the point), the cosine of that angle represents how much force will transfer to the object ball. the cosine of 0' is 1 (or 100%), and the cosine of 90' is 0. now, a half-ball hit (45') results in about 70.7% transfer, leaving 29.3% (minus a little friction) with the cueball. so it's not a linear thing, mnShooter's post is wrong.

now, on rails. they're all a little different. your guess is as good as mine :)

-s
Click to expand...
Sorry, both of you are wrong. :D

Steev, you have your angles mixed up. A half ball hit is a 30 degree angle, not a 45 degree angle. Draw it out and you'll realize your error.

mnShooter is correct, only if you're talking about velocity. Since the original question is about "energy", than you have to take the square root of your answer, since kinetic energy is (1/2)*M*V^2. If you lose half of the velocity, then you lose three-quarters of the energy and only one quarter of the original energy is maintained in the CB.

Just in the details. ;)
 
Last edited:
sigh. i type faster than i think. anyway, the cosine thing is true if you have the angle right ;)

and although he said energy, i bet he meant speed.

-s
 
I think he meant velocity, not energy (since velodity is what you really care about in a position route). The cosine thing is correct, but you have to make sure you're using cue ball tangent angle as the angle, as opposed to any other angle. Take the angle between the CB's initial direction (off the cue-stick), and the cue ball's tangent direction off the OB, and take the cosine of that. The cosine is the fraction of its speed it will have left.

Draw and follow will both change this speed. Follow will accelerate the CB in the direction of its original movement and draw will accelerate the CB opposite this direction. Since the cue ball will always contain some component of its original momentum except in totally full-ball hits, this means draw will decrease overall velocity after contact and follow will increase it (not instantaneously, but as the spin begins to "take").

As far as rail speed, that totally depends on running vs. check english. The cue ball can speed up after a rail or it can dramatically slow down, depending on english. There's no simple formula that could accurately model this, as far as I know. Also it depends greatly on the condition of the cushion rubber, the temperature of the cushion rubber, and the condition of the table (i.e. do the rails have dead spots where something isn't screwed in as tightly as it should be).

-Andrew
 
If we're only talking about velocity, then mnShooter is still correct...the fraction of velocity lost equals the fraction of OB hit. Here's my proof...

First you, want to obtain the cut angle for a given fraction of OB hit. To obtain this, you'd have to take the inverse sine function of 1 minus the fraction of OB hit...or arcsin (1-x), where is x the fraction of OB hit.

Then you want to find out the resulting velocity of the CB after impact, so you take the sine of the cut angle (not the cosine, because you want to find out the velocity of the CB, not OB, after impact). Then you have...

sin [arcsin (1-x)] = velocity of CB after impact.

Since sin and arcsin are inverse functions, they cancel out and all you have left is (1-x), which is the velocity of the CB after impact.

Now, the original question is how much velocity (okay, he said "energy", but let's stick with velocity) does the CB lose after impact. Therefore, you'd have to subtract this answer from 1 again, so you'd have 1 - (1-x), which turns out to just equal x. Ta da. The fraction of velocity lost equals the fraction of OB hit.

mnShooter was right all along...that is, if we're talking about velocity, not energy. :)
 
jsp said:
If we're only talking about velocity, then mnShooter is still correct...the fraction of velocity lost equals the fraction of OB hit. Here's my proof...

First you, want to obtain the cut angle for a given fraction of OB hit. To obtain this, you'd have to take the inverse sine function of 1 minus the fraction of OB hit...or arcsin (1-x), where is x the fraction of OB hit.

Then you want to find out the resulting velocity of the CB after impact, so you take the sine of the cut angle (not the cosine, because you want to find out the velocity of the CB, not OB, after impact). Then you have...

sin [arcsin (1-x)] = velocity of CB after impact.

Since sin and arcsin are inverse functions, they cancel out and all you have left is (1-x), which is the velocity of the CB after impact.

Now, the original question is how much velocity (okay, he said "energy", but let's stick with velocity) does the CB lose after impact. Therefore, you'd have to subtract this answer from 1 again, so you'd have 1 - (1-x), which turns out to just equal x. Ta da. The fraction of velocity lost equals the fraction of OB hit.

mnShooter was right all along...that is, if we're talking about velocity, not energy. :)
Click to expand...

Sheesh, the guy asked for a 'rule of thumb' ... you can't do trigonometry on your thumbs, only algebra !

Dave

Nice little proof btw, it's cool when those trig functions disappear.
 
Billy_Bob said:
Is there any kind of "rule" or rule of thumb for how much energy will be used up when making a hit? (How much energy CB will lose.)

Like the hit itself;
Full ball hit
3/4 ball hit
1/2 ball hit
1/4 ball hit
thin hit

And then rails;
hits one rail and loses x amount of energy (After contacting object ball).
hits two rails and loses x amount of energy.
hits three rails and loses x amount of energy.

So say I had a half ball hit and the cue ball was going to hit 3 rails before traveling to where I wanted it to go. Half ball hit would remove x amount of speed, 1st rail x, 2nd x, 3rd x, etc.
Click to expand...
Some posters are confused about energy and velocity. It is the energy, not the velocity that is proportional to how far the cue ball goes. A ball wth twice the energy will go twice as far (if rails aren't involved).

If you hit an object ball full with a rolling cue ball, the cue ball will follow about 1/6th as far as the object ball goes. If you hit an object ball half-ball with a rolling cue ball, the two balls will go about the same distance. A simple graphical way of showing this and working out all other cut angles is in articles at http://www.sfbilliards.com/articles/BD_articles.html

If you stun into a ball, the previous posters have more or less gotten the velocity part right. If the cue ball is originally moving at V, the object ball speed is V*cos(cut_angle) and the cue ball speed is V*sin(cut_angle). The ratio of distances travelled after the hit is the square of the tangent of the cut angle. However, as the balls acquire smooth rolling, their speeds drop to 5/7 of the original speeds.

Google will do the arithmetic for you. For example, if you shoot a half-ball shot with stun, the ratio of the distances travelled is:

tan(arcsin(1 - .5))^2 = 0.333333333

Just paste the tan(arcsin(1 - .5))^2 into the search window in google and it will give you the answer.

Note that a 45-degree cut with stun will give you equal distances travelled after the hit for the two balls. Most people wouldn't guess this. This is shown by: tan(45 degrees)^2 (try it in google).

As was shown by experiments described in those articles mentioned above, a ball going directly into a cushion loses 1/2 its speed (pretty nearly) or 3/4 of its energy. This assumes a rolling ball. If you hit a cushion at an angle or with side spin, things change.
 
mnShooter said:
Full ball hit - 100%
3/4 ball hit - 75%
1/2 ball hit - 50%
1/4 ball hit - 25%
thin hit - 0-10%

one rail - about 40% depends on the rail
Click to expand...

Thanks, this is what I was looking for. Just sort of a general "baseline" before shooting a shot.

Like... Ok this this is going to be a 1/2 ball hit, then it is going to need to hit 3 rails after the shot to get to where I need it for the next shot. Then I will have something like the above to think about before shooting the shot.

Then after I get this "baseline" down, I can factor in other things like running english, reverse english, etc.

But for now I'm learning to use multiple rails to get position on my next shot (instead of say drawing back) and using the correct speed is difficult when CB is hitting several rails after the shot. (for me)
 
I don't get into all the math and crap, because as soon as you apply any english to the CB...every thing changes...


But a rule of thumb that I follow is that a 1/2 ball hit, the CB and OB will travel approx. the same distance...after impact

the thinner you hit on the OB the further the CB will travel in relation to the OB's distance...

same type of effects on angles....

put two OBs on the table and one CB...practice hitting one OB and carom into the other....no rail, one rail , two rail, whatever it takes...real good practice for CB control and safety play.
 
BRKNRUN said:
...But a rule of thumb that I follow is that a 1/2 ball hit, the CB and OB will travel approx. the same distance...after impact...
Click to expand...

I've got that down! The 1/2 ball hit comes in quite handy when it is "snooker your opponent time". (Quite hard to learn speed of shot wise though.) As always, practice makes perfect.
 
jsp is correct... 1/2 ball hit is actually a 30 degree impact, 1/4 ball hit is a 50 degree impact... and to get a good 45 degree angle (for equal travel on the cue and object after impact), you actually have to hit about 1/3 of the object ball. This is all according to my handy-dandy CAD program....
 
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