Why CTE is silly

[Patrick Johnson]

Not this trip. But there will come a time when you and I will tangle again and you can deprive me of some more meals.

However if Pat will play me some one hole then I cover all the action you feel like stacking up. ;-)

JUST KIDDING!!! I don't want this to turn into another bark fest. Pat and I are going to have fun.

What is this "one hole" of which you speak? If it's not too demanding I might give it a whirl.

pj <- open to new things
chgo

 

[BRKNRUN]

Pay attention Pat: there is no spoon.

I suppose you are referrring to this....

This one sums it up pretty well....watch it...(it has a twist)


http://www.youtube.com/watch?v=3Bm-O4fNQ1w&feature=related

 

[JB Cases]

Again
First off im not promoting anyone,to each there own.


This is not how i normally aim but i do use it on certain type of shot's.
In this video i use pocket tightners which make the pockets 3"s which is tight.I do miss but hell we all do.
How ever you pocket ball thats great im not trying to say the pivot system
is the best but to say its Bs well that's nonsense.


Sorry for the video.
Not trying to piss people off with them so dont get angry.
Just having fun thats all.http://www.youtube.com/watch?v=in1BuGYMflM

Aw snap.

Didja have to go and burst so many bubbles at once?

 

[dr_dave]

What I don't understand is for all the brain power between you and Dave Alciatore and Bob Jewett, that you couldn't get Hal on a pool table with a video camera and work it out.

By the time you were looking up threads on RSB to learn who Hal was, I had been involved in a lot of discussion about aiming and had already twice talked to Hal Houle in person around a pool table. In fact one time he, Bob Jewett, and I were together at Family Billiards in San Francisco. You and some others seem to think anybody who doesn't embrace your enthusiasm must be not trying or unwilling to consider something new.

I have also spent countless hours and many years trying to understand and help other people better understand what CTE is. I have spoken to Jal on the phone several times, at great length. I have also communicated (by phone and e-mail) with Ron, Spidey, Stan, and other CTE/pivot-based proponents over the years. I have also read and participated in countless threads in online forums dealing with aiming systems, especially those involving CTE. I have also studied Spidey's blog page in detail multiple times. I have also carefully tried out at the table everything that has been suggested by the people, threads, and resources mentioned above.

I have certainly dedicate lots of time and effort to try to "work out" the CTE approach. However, I honestly don't think any amount of "brain power" will help explain CTE any better than is already explained here:

http://billiards.colostate.edu/threads/aiming.html#CTE​

I know CTE "works" for some people, but I think the primary reasons it does are listed here:

http://billiards.colostate.edu/threads/aiming.html#benefits​

Regards,
Dave

 

[lfigueroa]

So I beat the guy who beat Nevel? Awesome. That makes me feel even better about it. Normally I wouldn't care so much but it just so happens that my one victory over you came jsut after you and Deno Andrews had been hazing me pretty good about my adoption of Hal's system.

I beat Deno as well so it was a double header.

It doesn't take much to keep me happy that's for sure. Learning a great aiming system or two does the trick and beating a major detractor at his game using that system is icing on the cake.

However I should balance it out by saying that I did lose to Pat Johnson that same trip and Mike Page has relived me of lunch money on several occasions. So you are kind of the odd man out here.

Pat is going to get a chance to extend his lead here in a few weeks. I am hard in training going over reams of old RSB threads to refresh my self on the aiming wars and get pumped up.

So I am glad you blew off Hal. If you hadn't who knows, maybe you would have beaten solidly and snapped off more champions than just Larry? The world will never know now.

Even though your happiness comes at my expense, I'm good with it.

In fact, I don't think I've felt this way about another person's victory since I was a volunteer for The Special Olympics.

Enjoy, John

Lou Figueroa

 

[dr_dave]

I know that Mike could help, but just in case he doesnt (and assuming I understand your question based on skimming just a couple of your comments to Dr Dave)...

Some years ago Mike gave an analogy on RSB that explained clearly to me what happens when a ball is rolling with sidespin. I've used it many times since (thanks, Mike) to explain it to others. Here it is:

Imagine that there's a cylinder (like a barrel) exactly the right size so that it's just contained within the ball. The cylinder/barrel's top and bottom edges meet the ball's surface like two latitude lines drawn around a globe in the northern and southern hemispheres.

Now imagine that you've tilted the barrel and are rolling it along the table's surface like you'd roll a barrel on its edge. You can see that the axis of rotation is tilted, and the barrel's bottom edge rolls along the table in a straight line without sliding at all.

Now imagine the ball surrounding the barrel rolling along with it across the table. This is how a ball rolls across the table with sidespin - no sliding, just rolling. If you were to put wet ink on the table it would mark a line around the ball at its "southern latitude" line.

The remaining wrinkle is that as the ball rolls its sidespin wears off and its axis of rotation becomes more and more horizontal. To visualize this, imagine that the tilted barrel within the ball tilts more and more onto its side, and simultaneously becomes fatter and fatter and shorter and shorter, but it continues rolling on its edge (you should be able to visualize doing this with a barrel that changes this way as you roll it). Eventually, when all the sidespin wears off, the fatter, shorter barrel becomes a flat disc rolling vertically on its edge. The wet ink line marked on the ball is actually a spiral that starts as a small tilted latitude and becomes a vertical "great circle" longitude when the ball is rolling normally without sidespin.

Wow! I've just sat here for a good fifteen minutes thinking this through with the cylinder example. I see it now--even though it's counter-intuitive. A ball can roll in a STRAIGHT LINE on a non-equatorial axis**. Balls moving very slowly while they're spinning like hell on an only slightly tilted axis are actually translating that fast spin into a slow straight line by simple rolling. Whew!

Thank you very much, PJ. Dr. Dave: sorry for the argument (although, damn you, you COULD have mentioned the cylinder concept--which definitely allows for the right thought process and questions to be considered). I'm a "believer" now (but not in the sense that CTE users are "believers" ).

EDIT: ** I should say a "non-circumferential axis" to be clearer.

GetMeThere, I'm glad we finally reached "common ground" on the "roll" vs. "slide" discussion.

PJ, thanks for posting the excellent quote from Mike. I've added it to my spin/slide/roll FAQ page, which now has links to related resources on the topic, including an excellent illustration you posted in the past.

Regards,
Dave

 

[Patrick Johnson]

GetMeThere, I'm glad we finally reached "common ground" on the "roll" vs. "slide" discussion.

PJ, thanks for posting the excellent quote from Mike. I've added it to my spin/slide/roll FAQ page, which now has links to related resources on the topic, including an excellent illustration you posted in the past.

Regards,
Dave

It's not a direct quote from Mike - that was many years ago and he was probably more lucid than I am now. Mike's the patron saint of aha!.

pj
chgo

 

[dr_dave]

I tried both CTE today, and it works!

I tried both DAM and CTE today. DAM worked much better for most shots! Although, I was able to make CTE work if I used my DAM intuition/feel/aim during the "pivot."

 

[dr_dave]

If anyone is using 90/90 now and would like a diagram describing how it works, shoot me a pm. Note: this is free.

FYI, I have some diagrams (from cleary) illustrating 90/90 here:

http://billiards.colostate.edu/threads/aiming.html#ninety​

I would like to see what you have. Also, if you want to post it here or e-mail it to me, I would be happy to add it to my resource page (with credit to you).

Regards,
Dave

 

[Patrick Johnson]

PJ, thanks for posting the excellent quote from Mike. I've added it to my spin/slide/roll FAQ page, which now has links to related resources on the topic, including an excellent illustration you posted in the past.

Regards,
Dave

Dave, you have a similar description there to mine/Mike's, which helps me frame a question I've been meaning to ask the Collective Scientific Mind. Here's the description from your page:

If a ball were rolling on an inked surface, the ink trace on the ball would be a circle. If the ball is rolling with no sidespin, the circle is vertical and goes around the full circumference ("great circle") of the ball. If the ball is rolling with sidespin, the circle is smaller and tilted. With more sidespin and/or less forward speed, the circle is smaller and more tilted. A ball with lots of spin and very little forward roll would trace a very small circle. A ball spinning in place traces a point (the smallest possible circle). Every "rolling" ball traces a circle, and as the sidespin wears off, the circle size and tilt angle change.

Here's my question: Is the ratio of spin to speed directly represented by the tilt of the spin axis (i.e., how much of the vertical-to-horizontal change has taken place)? Does the spin/speed ratio change linearly with the tilt change? (I'm assuming spin/speed ratio means rotational RPMs-to-translational RPMs, and that they're equal when the axis is 45 degrees...?)

Thanks,

pj
chgo

 

[dr_dave]

GetMeThere, I'm glad we finally reached "common ground" on the "roll" vs. "slide" discussion.

PJ, thanks for posting the excellent quote from Mike. I've added it to my spin/slide/roll FAQ page, which now has links to related resources on the topic, including an excellent illustration you posted in the past.

It's not a direct quote from Mike - that was many years ago and he was probably more lucid than I am now.

I've changed the credit line to read: "from Patrick Johnson (paraphrasing Mike Page)." Please let me know if you think something else would be more appropriate.

Mike's the patron saint of aha!.

Agreed.

Regards,
Dave

 

[jake-a-licious]

Ok here i go i may get chided for this. I have never used cte. I have tried very many aiming systems. My point is this i didnt chart any numbers but, cte had me very confused from all the info i read about it until this morning. I had drawn countless pictures with my protractor and ruler of different cut angles and ob spots to hit. With how i understand cte is used all my drawing gave me enough proof to have faith in it with out using it yet. Im going to the pool hall today to go try it out. I dont have any way of showing you guys what i came up with cause i am not very computer savvy, sorry.

 

[dr_dave]

GetMeThere, I'm glad we finally reached "common ground" on the "roll" vs. "slide" discussion.

PJ, thanks for posting the excellent quote from Mike. I've added it to my spin/slide/roll FAQ page, which now has links to related resources on the topic, including an excellent illustration you posted in the past.

Dave, you have a similar description there to mine/Mike's, which helps me frame a question I've been meaning to ask the Collective Scientific Mind. Here's the description from your page:

If a ball were rolling on an inked surface, the ink trace on the ball would be a circle. If the ball is rolling with no sidespin, the circle is vertical and goes around the full circumference ("great circle") of the ball. If the ball is rolling with sidespin, the circle is smaller and tilted. With more sidespin and/or less forward speed, the circle is smaller and more tilted. A ball with lots of spin and very little forward roll would trace a very small circle. A ball spinning in place traces a point (the smallest possible circle). Every "rolling" ball traces a circle, and as the sidespin wears off, the circle size and tilt angle change.

Here's my question: Is the ratio of spin to speed directly represented by the tilt of the spin axis (i.e., how much of the vertical-to-horizontal change has taken place)? Does the spin/speed ratio change linearly with the tilt change? (I'm assuming spin/speed ratio means rotational RPMs-to-translational RPMs, and that they're equal when the axis is 45 degrees...?)

The "inked circle" (AKA, the bottom rim of the rolling barrel) is in a plane perpendicular to the spin axis. The spin axis direction is determined from the vector sum of the roll and sidespin angular velocity vectors. The axis of rotation for the roll (topspin) vector is horizontal, and the axis of rotation for sidespin is vertical. The vector combination of these two components gives the total spin rate and the resultant axis of rotation of the ball (AKA, the barrel axis). The forward speed of the ball is:

v = w*r​

where w is the total spin rate, and r is the perpendicular distance from the contact point (between the ball and the cloth) to the resultant spin axis, given by:

r = R*sin(A)​

where R is the radius of the ball and A is the angle of the tilt axis from the vertical. So the speed of the ball is related to the spin rate and axis tilt angle according to:

v = w*R*sin(A)​

For a ball rolling forward with no sidespin (i.e., horizontal spin axis: A=90 deg), v=wR; and for a ball spinning in place (i.e., vertical spin axis: A=0 deg), v=0.

The resultant spin is related to the roll spin rate (wr) and sidespin rate (ws) according to:

w = sqrt(wr^2 + ws^2)​

and the tilt axis is given by:

A = atan(wr/ws)​

I think what you are asking about is the spin ratio (SR) of dimensionless translational speed (which is the roll rate wr) to the sidespin rotational speed (ws):

SR = wr/ws = tan(A)​

As you point out, SR=1 when wr=ws, where the tilt axis is 45 deg. However, this ratio does not vary linearly with tilt angle.

Sorry about all of the math, but that seemed to be the best way to answer your questions clearly.

Regards,
Dave

 

[8pack]

deleteddddddddd

 

[randyg]

If i had to draw up cte it might look like this! (maybe?)

That's it......SPF=randyg

 

[sfleinen]

If i had to draw up cte it might look like this! (maybe?)

That kinda looks like one zoomed-in on an old weather vane on the top of a barn, going crazy in a wind storm.

 

[Patrick Johnson]

The "inked circle" (AKA, the bottom rim of the rolling barrel) is in a plane perpendicular to the spin axis. The spin axis direction is determined from the vector sum of the roll and sidespin angular velocity vectors. The axis of rotation for the roll (topspin) vector is horizontal, and the axis of rotation for sidespin is vertical. The vector combination of these two components gives the total spin rate and the resultant axis of rotation of the ball (AKA, the barrel axis). The forward speed of the ball is:

v = w*r​

where w is the total spin rate, and r is the perpendicular distance from the contact point (between the ball and the cloth) to the resultant spin axis, given by:

r = R*sin(A)​

where R is the radius of the ball and A is the angle of the tilt axis from the vertical. So the speed of the ball is related to the spin rate and axis tilt angle according to:

v = w*R*sin(A)​

For a ball rolling forward with no sidespin (i.e., horizontal spin axis: A=90 deg), v=wR; and for a ball spinning in place (i.e., vertical spin axis: A=0 deg), v=0.

The resultant spin is related to the roll spin rate (wr) and sidespin rate (ws) according to:

w = sqrt(wr^2 + ws^2)​

and the tilt axis is given by:

A = atan(wr/ws)​

I think what you are asking about is the spin ratio (SR) of dimensionless translational speed (which is the roll rate wr) to the sidespin rotational speed (ws):

SR = wr/ws = tan(A)​

As you point out, SR=1 when wr=ws, where the tilt axis is 45 deg. However, this ratio does not vary linearly with tilt angle.

Sorry about all of the math, but that seemed to be the best way to answer your questions clearly.

Regards,
Dave

Thanks, Dave.

pj
chgo

 

[ThePoliteSniper]

I tried both DAM and CTE today. DAM worked much better for most shots! Although, I was able to make CTE work if I used my DAM intuition/feel/aim during the "pivot."

Well, what I did wasn't exactly textbook CTE. I had problems aiming at the edge of the ob, so I started aiming the cuball to the desired contact point of the ob. You could call it CTCP. Then I realized, when aiming at the desired contact point there was no need for a pivot any more. That made it much easier for me. :smile-us-down:

 

[dr_dave]

I tried both DAM and CTE today. DAM worked much better for most shots! Although, I was able to make CTE work if I used my DAM intuition/feel/aim during the "pivot."

Well, what I did wasn't exactly textbook CTE. I had problems aiming at the edge of the ob, so I started aiming the cuball to the desired contact point of the ob. You could call it CTCP. Then I realized, when aiming at the desired contact point there was no need for a pivot any more. That made it much easier for me. :smile-us-down:

Sounds DAM good to me. :grin-square:

Regards,
Dave

 

[LAMas]

:smile::thumbup:

Well, what I did wasn't exactly textbook CTE. I had problems aiming at the edge of the ob, so I started aiming the cuball to the desired contact point of the ob. You could call it CTCP. Then I realized, when aiming at the desired contact point there was no need for a pivot any more. That made it much easier for me.

Are you aiming the center of the CB to the contact point on the OB? I tried that years ago and I undercut all of my shots except for straight in ones.

I wish it were that easy.:smile::thumbup: