So what happens when a ball burns up?

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So what happens when a ball burns up?

Post by deanchamp » August 10th, 2015, 11:33 pm

Hi all, I have been thinking about this a bit lately and was wondering about what actually happens to the ball as it loses energy, and why this happens.

Is it the strength of the coverstock against the friction of the lane? And I guess it is going to happen more when a ball flares early in the friction exposing fresh cover to the friction each rotation? So will the early friction make the ball flare less as it loses energy faster? And is this what kills the backend?

I have probably answered my own question a bit but I would like to be able to explain this to people in more detail as it is one of the most confusing things for bowlers to understand these days. Strong balls with strong layouts on a THS = no hook?
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Re: So what happens when a ball burns up?

Post by MWhite » August 11th, 2015, 12:56 am

deanchamp wrote:Hi all, I have been thinking about this a bit lately and was wondering about what actually happens to the ball as it loses energy, and why this happens.

Is it the strength of the coverstock against the friction of the lane? And I guess it is going to happen more when a ball flares early in the friction exposing fresh cover to the friction each rotation? So will the early friction make the ball flare less as it loses energy faster? And is this what kills the backend?

I have probably answered my own question a bit but I would like to be able to explain this to people in more detail as it is one of the most confusing things for bowlers to understand these days. Strong balls with strong layouts on a THS = no hook?

"Burned up" really means a loss of axis rotation.

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Re: So what happens when a ball burns up?

Post by stevespo » August 11th, 2015, 6:26 pm

MWhite wrote:"Burned up" really means a loss of axis rotation.
Yes, but to me it also means early loss of rotational/translational energy as well.

My thinking might be wrong (or simplistic) here, but you roll the ball and give it forward (translational) motion with your arm swing. You give it rotational energy (revolutions) with your hand/release. For a "typical" bowler that might mean 15 MPH forward (22 FPS) and 240 RPM (4 rev/sec, roughly 8 FPS). There is also axis tilt and axis of rotation, which influences motion and how the ball reacts. A "balanced" bowler will have roughly a 4/10 ratio between rotational and translational velocity, but tilt plays a role as it changes the track circumference.

During the skid phase (assuming there is oil in the front) the translational energy dominates. The ball goes fairly straight, but track flare, tilt and rotation also are at work. As the ball encounters friction, the rotational energy begins to exert influence and the ball starts to hook. The ball begins turning right (it loses axis of rotation), but moving left as it hooks (all assuming for a righty). It begins to equilibrate. When the two forces equalize, the ball rolls straight at whatever vector finally worked itself out.

If the ball finds friction too early, either because the head oil is toast or because you missed right on a house shot - you start losing energy (rotational and translational) and you also lose axis tilt and rotation too soon. This is what I would call "burning up" or "rolling out". Maybe they're describing slightly different things, but I believe it's fundamentally about losing too much energy early. The symptoms are painfully familiar - weak 10, pocket 7/10, possibly 2/8/bucket combinations.

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Re: So what happens when a ball burns up?

Post by Mo Pinel » August 12th, 2015, 12:40 am


A ball "burns up" when it enters the roll phase too soon. When the axis rotation and the axis tilt are equal (the 2nd transition), the ball will no longer hook. If the ball still has a little axis rotation and tilt (which are then equal), it will still hit. The ball will only stop hitting when both the axis rotation and tilt equal zero (true "roll out").

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Re: So what happens when a ball burns up?

Post by MWhite » August 12th, 2015, 3:49 am

Mo Pinel wrote:
A ball "burns up" when it enters the roll phase too soon. When the axis rotation and the axis tilt are equal (the 2nd transition), the ball will no longer hook.
That has been disputed.

The ball with at least some rev rate will change direction (still hook) if axis rotation is not perpendicular to the path of the ball, the ball is encountering at least some amount of friction, and the axis tilt is not 90 degrees from horizontal.

Newtons 3 laws apply to a bowling ball.

They aren't suspended just because the axis rotation, and axis tilt equal to each other.

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Re: So what happens when a ball burns up?

Post by MWhite » August 12th, 2015, 4:05 am

stevespo wrote:
Yes, but to me it also means early loss of rotational/translational energy as well.

My thinking might be wrong (or simplistic) here, but you roll the ball and give it forward (translational) motion with your arm swing. You give it rotational energy (revolutions) with your hand/release. For a "typical" bowler that might mean 15 MPH forward (22 FPS) and 240 RPM (4 rev/sec, roughly 8 FPS). There is also axis tilt and axis of rotation, which influences motion and how the ball reacts. A "balanced" bowler will have roughly a 4/10 ratio between rotational and translational velocity, but tilt plays a role as it changes the track circumference.

During the skid phase (assuming there is oil in the front) the translational energy dominates. The ball goes fairly straight, but track flare, tilt and rotation also are at work. As the ball encounters friction, the rotational energy begins to exert influence and the ball starts to hook. The ball begins turning right (it loses axis of rotation), but moving left as it hooks (all assuming for a righty). It begins to equilibrate. When the two forces equalize, the ball rolls straight at whatever vector finally worked itself out.

If the ball finds friction too early, either because the head oil is toast or because you missed right on a house shot - you start losing energy (rotational and translational) and you also lose axis tilt and rotation too soon. This is what I would call "burning up" or "rolling out". Maybe they're describing slightly different things, but I believe it's fundamentally about losing too much energy early. The symptoms are painfully familiar - weak 10, pocket 7/10, possibly 2/8/bucket combinations.

Steve
What a ball loses in translational energy, it gains in rotational energy, minus a small amount of energy converted to sound, and heat.

The ball will lose translational energy, and gain rotational energy until the rev rate (modified by tilt) makes the surface of the ball (at the point of contact with the lane) move at the same velocity as the center of the ball.

What is important is how much momentum the resulting shot has, what direction it ends up pointed towards, and at what location it hits the head pin.

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Re: So what happens when a ball burns up?

Post by stevespo » August 12th, 2015, 4:47 am

That makes sense to me. My physics is rusty and I'm just trying to improve my mental model. Mass is constant, so less energy means decreased velocity, with generally less angle, weaker entry and poor carry.

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Re: So what happens when a ball burns up?

Post by MWhite » August 12th, 2015, 5:46 am

stevespo wrote:That makes sense to me. My physics is rusty and I'm just trying to improve my mental model. Mass is constant, so less energy means decreased velocity, with generally less angle, weaker entry and poor carry.

Steve

Less translational energy compared to what it had at release.

However, compare two shots where one enters the dry area early and "burns up", and the other enters the dry area late enough that it reaches the roll phase on a line towards the pocket, with a strong angle of entry.


Both balls will slow down the same amount, and rev up the same amount.

So they both have the same velocity entering the pins.

The difference is where they hit the head pin, and at what angle.

A while back, LabRat did some calculations to determine the final velocity / rev rate based on a given starting velocity / rev rate using conservation of energy.

How far down the lane the the ball entered the roll phase, was not a determining factor.

Most balls that "burn up" tend not to hit the pocket, if they do, they hit with a shallow angle of entry.

That doesn't mean they can't carry, just less likely.

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Re: So what happens when a ball burns up?

Post by Mo Pinel » August 12th, 2015, 7:07 am

Well! That elicited many responses!!!!!! Good!!!!!! Some dissension is always good! Nice guys!

The result is that for every inch of lane covered, the ball track covers 1" on the surface of the ball. True pure roll! No skid factor! That's the important factor.

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Re: So what happens when a ball burns up?

Post by LabRat » August 12th, 2015, 10:27 am

MWhite wrote: What a ball loses in translational energy, it gains in rotational energy, minus a small amount of energy converted to sound, and heat.

The ball will lose translational energy, and gain rotational energy until the rev rate (modified by tilt) makes the surface of the ball (at the point of contact with the lane) move at the same velocity as the center of the ball.

What is important is how much momentum the resulting shot has, what direction it ends up pointed towards, and at what location it hits the head pin.
While this is true for the average speed of the ball track, when a ball rolls the contact point on the lane is not moving with respect to the lane. This is why a rolling ball hits better - the coefficient of static friction is almost always highr than sliding friction, so a rolling ball grips the lane better. It always harder to gets something sliding from rest than to keep it moving once it starts.
I don't agree with the 'rolls when axis tilt = rotation' dogma, for reasons previously posted. The physics simply says it's wrong.
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Re: So what happens when a ball burns up?

Post by ads » August 12th, 2015, 2:50 pm

MWhite wrote:

Less translational energy compared to what it had at release.

However, compare two shots where one enters the dry area early and "burns up", and the other enters the dry area late enough that it reaches the roll phase on a line towards the pocket, with a strong angle of entry.


Both balls will slow down the same amount, and rev up the same amount.

So they both have the same velocity entering the pins.

The difference is where they hit the head pin, and at what angle.

A while back, LabRat did some calculations to determine the final velocity / rev rate based on a given starting velocity / rev rate using conservation of energy.

How far down the lane the the ball entered the roll phase, was not a determining factor.

Most balls that "burn up" tend not to hit the pocket, if they do, they hit with a shallow angle of entry.

That doesn't mean they can't carry, just less likely.
2 questions (sorry, no physics background).

1. Will ball travel faster on oil (skid) and slower on dry (friction)? If then, why 2 balls will slow down the same amount and rev up the same amount when 1 enters the dry zone earlier (leaves the oil earlier as well).

2. What are the characteristics of roll phase if not AR=AT?
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Re: So what happens when a ball burns up?

Post by Mo Pinel » August 12th, 2015, 2:57 pm

ads wrote:
2 questions (sorry, no physics background).

1. Will ball travel faster on oil (skid) and slower on dry (friction)? If then, why 2 balls will slow down the same amount and rev up the same amount when 1 enters the dry zone earlier (leaves the oil earlier as well).

2. What are the characteristics of roll phase if not AR=AT?

One thing that is true and not open to conjecture and has been measured is that the path of the ball is a straight line in the roll zone (linear equation). That's what's important to ball motion.

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Re: So what happens when a ball burns up?

Post by stevespo » August 12th, 2015, 3:33 pm

MWhite wrote: However, compare two shots where one enters the dry area early and "burns up", and the other enters the dry area late enough that it reaches the roll phase on a line towards the pocket, with a strong angle of entry.

Both balls will slow down the same amount, and rev up the same amount.
I can't dispute or disprove this first statement, and if LabRat has done the math it's that much more convincing - but it certainly appears that the ball entering the friction "too early" is rolling more slowly when it hits the pins. Perhaps it's merely perception, due to less angle or some other factor (like actually throwing it more slowly on the miss right).

Sure would love to see some CATS data, represented graphically - to help understand this.

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Re: So what happens when a ball burns up?

Post by MegaMav » August 12th, 2015, 3:45 pm

I've never seen "burn up" in my life as an observer and bowler.
Things I have seen, accused of being "burn up":

1. Early hook, and less angular motion on the backend.
2. Poor layouts usually relating to too long of a Pin to PAP on asymmetrics for low lilt bowlers causing premature ending of the hook phase before and acceptable entry angle.
3. Balls going too long from not enough surface friction and too high of angle sums to recover.

I have heard comments about "burn up" but I have never understood what they're talking about, and I think its used as a catch all phrase by bowlers that dont know what they're looking at or dont know what they're talking about.
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Re: So what happens when a ball burns up?

Post by JohnP » August 12th, 2015, 4:51 pm

If you want to see burn up ask your house manager if you can throw one ball on a lane that has been stripped but not oiled. -- JohnP

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Re: So what happens when a ball burns up?

Post by purduepaul » August 12th, 2015, 5:24 pm

Hello all,

Allow me to answer the actual questions and then I will address the AR=AT debate.

Flare's effect on friction is an inverse logarithmic scale, meaning at a certain amount of flare, the friction gained by additional flare in very minimal. It's probably somewhere between 4-5"
Early versus later friction doesn't influence friction that much. Flare is controlled more so by the core than the cover.

The second transition is defined as when axis rotation = axis tilt. The people who think otherwise and who are using newton's laws are not understanding the whole energy transfer on a bowling ball.

Once the bowler imparts energy to the bowling ball, rotational and translational energy are applied to a bowling ball. During the skid phase of ball motion, due to the oil on the lane a very small amount of translational energy is lost due to friction AND a small amount of energy is transferred from translational energy to rotational energy. In the hook phase of ball motion more friction is lost by translational energy and a good amount of energy is transferred from translational energy to rotational energy. At the second transition point which is when ar=at translational energy stops being transferred to rotational energy. Translational energy is still lost due to friction. So by definition the most energy transferred from the ball to the pins will occur when the pins are impacted right after the second transition point.

If you have any other questions feel free to ask.

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Re: So what happens when a ball burns up?

Post by MWhite » August 12th, 2015, 7:54 pm

Mo Pinel wrote:
One thing that is true and not open to conjecture and has been measured is that the path of the ball is a straight line in the roll zone (linear equation). That's what's important to ball motion.

With the possible exception of this if they were to do away with static weight rules:

By flipping the weights, the ball could continue to curve to the left after the ball has achieved the roll phase.

[youtube][/youtube]

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Re: So what happens when a ball burns up?

Post by Mo Pinel » August 12th, 2015, 8:30 pm

MWhite wrote:

With the possible exception of this if they were to do away with static weight rules:

By flipping the weights, the ball could continue to curve to the left after the ball has achieved the roll phase.

[youtube][/youtube]
That video is from a very extreme static weight example. I call it the "light ball syndrome".

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Re: So what happens when a ball burns up?

Post by MWhite » August 13th, 2015, 3:24 am

stevespo wrote:
I can't dispute or disprove this first statement, and if LabRat has done the math it's that much more convincing - but it certainly appears that the ball entering the friction "too early" is rolling more slowly when it hits the pins. Perhaps it's merely perception, due to less angle or some other factor (like actually throwing it more slowly on the miss right).

Sure would love to see some CATS data, represented graphically - to help understand this.

Steve
The following is based on a few assumptions, both balls achieve the roll phase, both balls are released with the same specs (velocity, rev rate, tilt, axis rotation)

If one ball starts out at 17 mph, and skids 40 feet at that speed, then slows down 2 mph in the last 20 feet, it's impacting the pins at 15 mph.

A second ball starts out at 17 mph, but only skids 20 feet, slows down 2 mph in 20 feet, then rolls at 15 mph the final 20 feet, impacting the pins at 15 mph.

The first ball will have a higher average speed, but at impact, they are both going to be the same.

Once the ball is rolling, the friction that would reduce velocity, is insignificant.

Watch a youngster roll a ball down the lane, 1/2 way down you would swear it's about to stop, but it keeps on going. It's not very often you see a ball stop on the lane, unless it's been hitting the bumpers frequently.

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Re: So what happens when a ball burns up?

Post by LabRat » August 13th, 2015, 8:56 am

purduepaul wrote:Hello all,

Allow me to answer the actual questions and then I will address the AR=AT debate.

Flare's effect on friction is an inverse logarithmic scale, meaning at a certain amount of flare, the friction gained by additional flare in very minimal. It's probably somewhere between 4-5"
Early versus later friction doesn't influence friction that much. Flare is controlled more so by the core than the cover.

The second transition is defined as when axis rotation = axis tilt. The people who think otherwise and who are using newton's laws are not understanding the whole energy transfer on a bowling ball.

Interesting. I guess you can define the second transition in any way you want. Just to be clear, does your definition above also correspond (in your opinion) to the point at which the ball transitions from hook to pure roll (where the contact point is stationary wrt the lane surface)?
In terms of Newtons' Laws of Motion, I hope you mean that you believe that people are misapplying them rather than implying they do not apply to bowling ball motion.
I have stated the reasons why I believe that at=ar is not required for the ball to roll in other threads, even though for many bowlers it may be approximately correlated at the roll point. Using Blueprints' definition of the point where axis rotation decays to 0 wrt the direction of travel of the ball is independent of external factors such as launch angle and is I believe the best definition of the second transition.


Once the bowler imparts energy to the bowling ball, rotational and translational energy are applied to a bowling ball. During the skid phase of ball motion, due to the oil on the lane a very small amount of translational energy is lost due to friction AND a small amount of energy is transferred from translational energy to rotational energy. In the hook phase of ball motion more friction is lost by translational energy and a good amount of energy is transferred from translational energy to rotational energy. At the second transition point which is when ar=at translational energy stops being transferred to rotational energy. Translational energy is still lost due to friction. So by definition the most energy transferred from the ball to the pins will occur when the pins are impacted right after the second transition point.

Nitpick perhaps, but the most energy transferred to the pins would happen if the ball skidded all the way, losing a minimum of energy to friction. However, the loss of entry angle and the fact that the ball is still skidding makes this a less effective delivery. Deflection is minimised when the ball is rolling, not skidding, because static friction is greater than sliding, and the importance of entry angle is obvious. This allows the ball to drive through the pins, rather than just into them.

If you have any other questions feel free to ask.

Paul Ridenour
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Engineering consultant
Comments above.
With respect to burning up, I think that a ball that transitions more slowly will be more efficient at transferring translation energy to rotational energy, meaning less energy lost to friction. Abrupt changes in friction are going to mean more energy transferred to heat rather than rotation. You lose a lot more rubber doing burnouts than laps.
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