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Author Topic: Release Angle experiment  (Read 6692 times)

Gizmo823

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Release Angle experiment
« on: October 25, 2013, 12:45:12 PM »
Ok, I want to try to generate some layouts with this method.  All I need to know is your angle of rotation, and any angle you play on any shot.  Could be an angle you play on a house shot, broken down house shot, Shark pattern, Main Street, Stone Street, US Open, don't care.  Just think of the line you play and what angle you estimate it at, and I'll give you a dual angle sum to work from and you tell me if it's realistic.  Even if you change your angle of rotation on different shots, just say like the US Open, tell me what your angle of rotation is and what angle you throw at (if straight at the headpin from the dead center of the lane is 0, and from one side of the foul line to the other is 90).  I'm really hoping I can generate a layout for you without needing to know anything else, that's the point. 
What would you be if you were attached to another object by an inclined plane, wrapped helically around an axis?

 

Aloarjr810

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Re: Release Angle experiment
« Reply #1 on: October 25, 2013, 01:19:11 PM »
okay on the THS here:

my Rotation: 65 degrees (estimated from video using the overlay graphic)

As for angle I'll say about 4 degrees

I stand about 25 and shoot about10-12 at the arrows

heres a video if it helps.
http://www.youtube.com/watch?feature=player_embedded&v=lc8wdUMmKyw
Aloarjr810
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Gizmo823

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Re: Release Angle experiment
« Reply #2 on: October 25, 2013, 01:47:56 PM »
Thanks!  I'd say the angle is closer to 10 degrees, but I'll calculate it both ways just for the hell of it.  So here's the deal, you base the ball choice on your speed, revs, and lane condition.  This definitely won't work if you pick a ball that's either too strong or too weak for you. 

Your rotation is 65 degrees, let's round your 4 degree angle to 5, for a total of 70.  Going into my calculator, that gets me to an ideal dual angle sum of 120.  That can be applied any way you want it, you could go 90-30, 60-60, etc.  The ratio and pin distance is up to you also depending on what shot shape you want.  Does a dual angle sum of 120 sound accurate or at the very least reasonable to work from? 

I've figured out it's definitely a "sweet spot" calculation, it doesn't apply if you go with a ball too weak or too strong for you or the condition.  If you're playing your A game, you pick out a benchmark ball, and I give you an angle sum of 120 to work from, is that anywhere close? 

I'm finding it hard to explain clearly, because I've heard "well what about condition?  ball speed?  rev rate?"  I'd factor all of that into ball choice and surface, because I really don't see the point of buying a ball that doesn't fit what you want and trying to drill it to fit you.  If you buy a ball that fits you and fits the condition, you SHOULD be able to use this method to get an ideal dual angle sum based on what angles you are playing or intend to play with the ball, because everybody has a reason for buying a ball, whether it's to play deeper, straighter, through carrydown, etc.  Considering that you pick the right ball, I haven't been able to poke a hole in this method yet, but lol that's why I'm asking people here . .

okay on the THS here:

my Rotation: 65 degrees (estimated from video using the overlay graphic)

As for angle I'll say about 4 degrees

I stand about 25 and shoot about10-12 at the arrows

heres a video if it helps.
http://www.youtube.com/watch?feature=player_embedded&v=lc8wdUMmKyw
What would you be if you were attached to another object by an inclined plane, wrapped helically around an axis?

Gizmo823

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Re: Release Angle experiment
« Reply #3 on: October 25, 2013, 01:55:46 PM »
And then again this whole thing could be overcomplicating something that's already involved, or it could just be completely superfluous. 

Oh and the 10 degree calculation would be 65+10 (your angle of rotation and laneplay or release angle) for 75.  Based off the mean or middle of the recommended dual angle range, which is 95, 95-75=20.  Now because 75 degrees isn't much of an angle for the ball to overcome, you need to go with a less aggressive layout, so if your total number is UNDER 95, you ADD the difference to 95, which in this case makes it 115.  If your angle is greater than 95, then you subtract the difference from 95 to get a more aggressive layout.  So for a 65 degree angle of rotation and a 10 degree release angle, my calculation would give you an ideal dual angle sum of 115, which again can be divvied up any way you want it. 
What would you be if you were attached to another object by an inclined plane, wrapped helically around an axis?

Aloarjr810

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Re: Release Angle experiment
« Reply #4 on: October 25, 2013, 02:38:38 PM »
Quote
Does a dual angle sum of 120 sound accurate or at the very least reasonable to work from

Well not a expert on the DA stuff. (hopefully you'll get some more knowledgeable people to test on)

But the 3 balls I have now the sum ranges from 125 to 135

90 X 4" X 40
90 X 4" X 45
90 X 4" X 35

So 120 seems close.

I was told my  Dual Angle Sweet Spot looked to be around Sweet Spot 90* +/-20,

I estimated it at a 100 +/-10

Aloarjr810
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Gizmo823

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Re: Release Angle experiment
« Reply #5 on: October 25, 2013, 02:43:29 PM »
Fair enough, thanks!

Quote
Does a dual angle sum of 120 sound accurate or at the very least reasonable to work from

Well not a expert on the DA stuff. (hopefully you'll get some more knowledgeable people to test on)

But the 3 balls I have now the sum ranges from 125 to 135

90 X 4" X 40
90 X 4" X 45
90 X 4" X 35

So 120 seems close.

I was told my  Dual Angle Sweet Spot looked to be around Sweet Spot 90* +/-20,

I estimated it at a 100 +/-10
What would you be if you were attached to another object by an inclined plane, wrapped helically around an axis?

itsallaboutme

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Re: Release Angle experiment
« Reply #6 on: October 25, 2013, 02:47:25 PM »
How are you calculating release angle?  Seems to me you are guessing at numbers.  For your whole system to work this needs to be a measurement and not a random number.  In your other thread you estimate you playing 20-7 as 45 degrees.  To play that you are probably setting the ball down around 27.  So your balls is 20 inches vertical in 42 or 43 feet horizontal.  Help a brotha out here.

Gizmo823

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Re: Release Angle experiment
« Reply #7 on: October 25, 2013, 02:57:32 PM »
Yeah, right now I'm just guessing and working out the concept before I actually try to do the math on working out the angle.  Lol you're not missing anything, I'm just apparently not that great at estimating angles on the lane . . and I also have to figure out at what distance to calculate the angle at.  20 at release to 7 at 35 feet would be a different angle than 20-7 at 40 or 45 feet.

How are you calculating release angle?  Seems to me you are guessing at numbers.  For your whole system to work this needs to be a measurement and not a random number.  In your other thread you estimate you playing 20-7 as 45 degrees.  To play that you are probably setting the ball down around 27.  So your balls is 20 inches vertical in 42 or 43 feet horizontal.  Help a brotha out here.
What would you be if you were attached to another object by an inclined plane, wrapped helically around an axis?

Gizmo823

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Re: Release Angle experiment
« Reply #8 on: October 25, 2013, 03:02:24 PM »
And it may or may not even be worth anything, I just saw what I thought could be a correlation and am working through the logistics, but I AM seeing now that the release angles are going to be a lot smaller than I thought they'd be.  Playing a 45 degree angle would be pretty ridiculous and unneccesary . .
What would you be if you were attached to another object by an inclined plane, wrapped helically around an axis?

Gizmo823

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Re: Release Angle experiment
« Reply #9 on: October 25, 2013, 03:05:26 PM »
But it is relevant in some way, if your angle of rotation is relevant, then the angle you play on the lane is relevant.  There just may not be as direct of a correlation as I'd thought at first.  It's seems to be similar to the dual angle sweet spot calculation, just not quite as involved, and seeing as how this only considers angles, I might have to do some experimenting there too. 
What would you be if you were attached to another object by an inclined plane, wrapped helically around an axis?

itsallaboutme

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Re: Release Angle experiment
« Reply #10 on: October 25, 2013, 03:30:53 PM »
At some point you hit the law of diminishing returns.  This stuff is WAY beyond that point. 

If you've already got 2 dogs take them for a walk.  Or teach them how to ride a skateboard. 

Gizmo823

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Re: Release Angle experiment
« Reply #11 on: October 25, 2013, 03:46:26 PM »
True . .

At some point you hit the law of diminishing returns.  This stuff is WAY beyond that point. 

If you've already got 2 dogs take them for a walk.  Or teach them how to ride a skateboard.
What would you be if you were attached to another object by an inclined plane, wrapped helically around an axis?

kidlost2000

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Re: Release Angle experiment
« Reply #12 on: October 25, 2013, 11:59:14 PM »
They are relavant but not with what you are trying to do. Release angle is key. Angle on the lane isn't a constant. All of the bowlers other stats plus the lane condition plus the ball would have to ba calculated into the equation.

Its a nice idea.
…… you can't  add a physics term to a bowling term and expect it to mean something.

LuckyLefty

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Re: Release Angle experiment
« Reply #13 on: October 26, 2013, 04:09:12 PM »
Release angle of rotation I get.

Straight up 9 on a shot does that equal a 0 initial angle of lane play?

THe shot I am playing at the safe zone for not hanging and leaving a washout is 12 to 9 up to 14 to 9.

Almost anything steeper on the left is a flat 7.  IS that the angle you are talking about? 

REgards,

Luckylefty
PS sorry for starting behind the rest of the discussion.  4 inch pin to pap in most cases is too much flare and midlane except for the weakest of cores at this house
It takes Courage to have Faith, and Faith to have Courage.

James M. McCurley, New Orleans, Louisiana

LuckyLefty

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Re: Release Angle experiment
« Reply #14 on: October 28, 2013, 06:07:05 AM »
To the Original poster Gizmo....please explain?

Regards,

Luckylefty
It takes Courage to have Faith, and Faith to have Courage.

James M. McCurley, New Orleans, Louisiana