# Can you reduce engine braking? 2008 yz 450f

59 replies to this topic
• # KJ790

Posted September 20, 2012 - 09:52 AM

Then there would be no motion. When the glue breaks, the forces are put out of balance with each other (one is reduced to near zero), and the centrifugal force wins, propelling the ball off the edge.

Not true, motion is not dependent on force. Force is defined by an acceleration. The ball was in motion when the glue was holding, but it was being redirected into a circular motion by the glue (which would be an acceleration, and hence centripetal force exists). Once the glue breaks the ball is still in motion, but there is no longer any force on it, so it will continue in motion in the direction that it was traveling when the glue broke, which would be away from the disk. If there were a force being applied to the ball then an acceleration would have to exist, however the ball does not change velocity or direction, so no force is applied to the ball as it flies away from the disk (other than wind resistance, but we can neglect that).

• # jaro51

Posted September 20, 2012 - 10:31 AM

Not true, motion is not dependent on force. Force is defined by an acceleration. The ball was in motion when the glue was holding, but it was being redirected into a circular motion by the glue (which would be an acceleration, and hence centripetal force exists). Once the glue breaks the ball is still in motion, but there is no longer any force on it, so it will continue in motion in the direction that it was traveling when the glue broke, which would be away from the disk. If there were a force being applied to the ball then an acceleration would have to exist, however the ball does not change velocity or direction, so no force is applied to the ball as it flies away from the disk (other than wind resistance, but we can neglect that).

I think there's a place on TT for this type of discussion, the "engineering & Technology" forum. Maybe the moderator could move the tech discussion there and keep this thread on point about BTL clutches and reducung engine breaking. Sarcasim asside, it's a very interesting debate and I hate to poke fun at Grayracer since he's a wealth of knowledge around here!

Edited by jaro51, September 20, 2012 - 10:33 AM.

• # grayracer513

Posted September 20, 2012 - 02:19 PM

Not true, motion is not dependent on force. Force is defined by an acceleration. The ball was in motion when the glue was holding, but it was being redirected into a circular motion by the glue (which would be an acceleration, and hence centripetal force exists). Once the glue breaks the ball is still in motion, but there is no longer any force on it, so it will continue in motion in the direction that it was traveling when the glue broke, which would be away from the disk. If there were a force being applied to the ball then an acceleration would have to exist, however the ball does not change velocity or direction, so no force is applied to the ball as it flies away from the disk (other than wind resistance, but we can neglect that).

I'm surprised, KJ. I know you know that force can and often does exist in the absence of motion. Twist a door knob that's locked and you have an example of torque (force) that moves nothing. Push a wall and you have another example of force without motion. Power depends on force and motion.

I think there's a place on TT for this type of discussion, the "engineering & Technology" forum. ... Sarcasim asside, it's a very interesting debate

Trouble is that in the E&T forum, people only want to post things that belong here instead.

• # KJ790

Posted September 20, 2012 - 03:25 PM

I'm surprised, KJ. I know you know that force can and often does exist in the absence of motion. Twist a door knob that's locked and you have an example of torque (force) that moves nothing. Push a wall and you have another example of force without motion. Power depends on force and motion.

Haha, this is getting fun, I love talking about stuff like this!

But force by definition is F=ma. If you push on a wall, the net force of the system is zero because you push on the wall, and the floor pushes on the wall in the opposite direction. In the same way the wall pushes back on you and the floor pushes on you in the opposite direction. No acceleration of the wall in respect to the house, therefore a net force of zero on the wall. Yes there are individual forces in that system, but they all have an equal and opposite force keeping the object static. As I said before, motion is not dependent on force, but acceleration is.

Now with the ball on the spinning disk example you said that centrifugal force pushes the ball away from the disk once the glue breaks, meaning that the net force on the ball would not be zero. Since F=ma, and the ball has a mass, there must be an acceleration of the ball if the forces are not equal. Say that the ball was traveling due west at 10 fps right at the instant that the glue let loose. What direction and what velocity would the ball be going a second later? Still due west and 10 fps (though it would now be 10ft away from the disk). The ball did not accelerate or change direction, so there was no force acting on it, only inertia.

Edited by KJ790, September 20, 2012 - 03:25 PM.

• # grayracer513

Posted September 20, 2012 - 04:02 PM

Say that the ball was traveling due west at 10 fps right at the instant that the glue let loose. What direction and what velocity would the ball be going a second later? Still due west and 10 fps (though it would now be 10ft away from the disk). The ball did not accelerate or change direction, so there was no force acting on it, only inertia.

Ah, but is that the case? A simple test: Take any object on a string and spin it in a circle and let's see what you have. There's the object being subjected to all this abuse, and if we look at things we see it acted on by 3 things, primarily: the froce trying to spin it in the circle, and the string, which becomes centripetal force as it is holding/pulling the object toward the center. This creates a lateral acceleration toward the circle vs. the kinetic tendency to continue straight on a tangent with the circle. That acceleration creates a centrifugal force that pushes out from the center, extending the string.

I know you already disagree with that because in the minds of degreed engineers, the force extending the string is the centripetal. But we're getting to the good part.

Now that you have it spinning, your position on this is that if the centripetal force is abruptly terminated, the object will continue in a straight line in the direction it's going at the time. So which direction is it going? If I am spinning this setup over my head and facing 12:00 o'clock with 3:00 being at my right, and spinning it counterclockwise, then at the moment it reaches 12:00, the object is going where? Well, it's going in a circle, of course, and not straight in any direction. But according to what you've said, if it were not for the centripetal force accelerating it toward the center (thereby making it turn), it would continue forward from wherever and not move toward the center. So if, at this point, I terminate the centripetal force by releasing the string, the centrifugal force should also be eliminated and the object will fly straight to 9:00 on my left because that's where it was going when I let go of it.

Why then does it instead fly to about 11:30 when released at 12:00, hmm? Because at the instant I released the string the centripetal force was cut off immediately while the centrifugal force built up in the unit's mass had not been relieved yet. That force then accelerated it outward away from the center.

• # KJ790

Posted September 20, 2012 - 04:27 PM

Ah, but is that the case? A simple test: Take any object on a string and spin it in a circle and let's see what you have. There's the object being subjected to all this abuse, and if we look at things we see it acted on by 3 things, primarily: the froce trying to spin it in the circle, and the string, which becomes centripetal force as it is holding/pulling the object toward the center. This creates a lateral acceleration toward the circle vs. the kinetic tendency to continue straight on a tangent with the circle. That acceleration creates a centrifugal force that pushes out from the center, extending the string.

I know you already disagree with that because in the minds of degreed engineers, the force extending the string is the centripetal. But we're getting to the good part.

Now that you have it spinning, your position on this is that if the centripetal force is abruptly terminated, the object will continue in a straight line in the direction it's going at the time. So which direction is it going? If I am spinning this setup over my head and facing 12:00 o'clock with 3:00 being at my right, and spinning it counterclockwise, then at the moment it reaches 12:00, the object is going where? Well, it's going in a circle, of course, and not straight in any direction. But according to what you've said, if it were not for the centripetal force accelerating it toward the center (thereby making it turn), it would continue forward from wherever and not move toward the center. So if, at this point, I terminate the centripetal force by releasing the string, the centrifugal force should also be eliminated and the object will fly straight to 9:00 on my left because that's where it was going when I let go of it.

Why then does it instead fly to about 11:30 when released at 12:00, hmm? Because at the instant I released the string the centripetal force was cut off immediately while the centrifugal force built up in the unit's mass had not been relieved yet. That force then accelerated it outward away from the center.

You lost me on this one. If you have the set up you are talking about and you let go of the string right when the object is at 12:00 then the object will fly directly left of the 12:00 point of your circle. It will not come back down towards 11:30 or 9:00 at all. If you took a snap shot of the object at 12:00 it's entire velocity would be directly to the left. If you let go of that string at that very point then the object's inertia will make it continue in that very direction, which would be directly left of the "top" of your clock.

• # FinchFan194

Posted September 20, 2012 - 06:49 PM

This is a dual of two of the smartest mo fros I know of on this forum!

• # grayracer513

Posted September 21, 2012 - 05:58 AM

If you have the set up you are talking about and you let go of the string right when the object is at 12:00 then the object will fly directly left of the 12:00 point of your circle. It will not come back down towards 11:30 or 9:00 at all.

Funny. 11:30 actually is just to the left of 12:00. 3:00 is at the right, remember?

I think you need to try this whole thing out for yourself. What you'll see is that when you release the string at 12:00, the object will end up, depending on its speed, say about 4 feet left of 12:00, but about 10 feet out from center. We both know why it goes to the left, but what you haven't explained is why it moves so far from center on a vector perpendicular to its forward travel around the arc.

Anything will work for the test, like a tennis ball in a grocery bag on a string, or maybe the mouse off of your computer.

• # KJ790

Posted September 21, 2012 - 06:15 AM

Funny. 11:30 actually is just to the left of 12:00. 3:00 is at the right, remember?

I think you need to try this whole thing out for yourself. What you'll see is that when you release the string at 12:00, the object will end up, depending on its speed, say about 4 feet left of 12:00, but about 10 feet out from center. We both know why it goes to the left, but what you haven't explained is why it moves so far from center on a vector perpendicular to its forward travel around the arc.

Anything will work for the test, like a tennis ball in a grocery bag on a string, or maybe the mouse off of your computer.

Think about the free-body diagram for the ball on the string. If the ball is spinning at a constant rate then the only force being applied to the ball is the tension on the string, redirecting it's path into a circular motion. This would be the centripetal force. The ball is applying an equal and opposite force back on the string, this would be the centrifugal force. This centrigugal force is not acting on the ball at all, it is acting on the string.

At any snap shot in time the ball's velocity is perpendicular to the string. So if the string breaks the ball centripetal force from the string goes to zero (and the centrifugal force on the string goes to zero). At this point in time the ball will continue on in the direction perpendicular to the string's position when it broke. As far as the ball is concerned, the instant the string breaks it is as if it was never traveling in a circle at all, but rather that it was traveling in a straight lin all along. The ball will not curve or arc at all. If ball was spinning counter-clockwise and the string breaks at 12:00, then the ball will fly away directly to the left, tangent to the original circular motion. If your string was 10ft long then the motion of the ball after the string broke would be the same as if you walked 10ft forward (so you were standing on the 12:00 mark rather than in the center of the circle), turned 90 degrees left, and threw the ball straight.

The ball will always fly in the direction of the V vector.

• # grayracer513

Posted September 21, 2012 - 06:25 AM

Nice theory. I understood this to be your position already. Now try it out and see what actually happens.

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• # KJ790

Posted September 21, 2012 - 06:36 AM

Nice theory. I understood this to be your position already. Now try it out and see what actually happens.

Haha, I have tried it out Gray, I really don't understand what you are trying to say. Every time I have tried it the ball goes perpendicular to the string, there is nowhere else it can go (other than to the ground from the force of gravity). If there is a force on the ball after the string breaks then it would have to accelerate in some way, unless you are suggesting that Newton's second law is incorrect (in which case everything we think we know about our world would be incorrect).

So if I understand you correctly, you are saying that after the string breaks the ball will continue to turn and travel in a direction other than the V vector at the point of the string breaking? What is applying this force to make it turn? Inertia is always in the direction of velocity, and there is nothing else in contact of the ball. There will be gravitational forces between the ball and your body, but this would be so small that they are negligible. There is nothing else acting on the ball after the string breaks.

If you want to throw a ball straight up you swing your arm out in front of you and let go of the ball when you arm is horizontal, the ball will go straight up in a path perpendicular to your arm. Same thing.

Edited by KJ790, September 21, 2012 - 06:36 AM.

• # bg10459

Posted September 21, 2012 - 07:00 AM

Think about the free-body diagram for the ball on the string. If the ball is spinning at a constant rate then the only force being applied to the ball is the tension on the string, redirecting it's path into a circular motion. This would be the centripetal force. The ball is applying an equal and opposite force back on the string, this would be the centrifugal force. This centrigugal force is not acting on the ball at all, it is acting on the string.

At any snap shot in time the ball's velocity is perpendicular to the string. So if the string breaks the ball centripetal force from the string goes to zero (and the centrifugal force on the string goes to zero). At this point in time the ball will continue on in the direction perpendicular to the string's position when it broke. As far as the ball is concerned, the instant the string breaks it is as if it was never traveling in a circle at all, but rather that it was traveling in a straight lin all along. The ball will not curve or arc at all. If ball was spinning counter-clockwise and the string breaks at 12:00, then the ball will fly away directly to the left, tangent to the original circular motion. If your string was 10ft long then the motion of the ball after the string broke would be the same as if you walked 10ft forward (so you were standing on the 12:00 mark rather than in the center of the circle), turned 90 degrees left, and threw the ball straight.

The ball will always fly in the direction of the V vector.

I'm not really sure how you can dispute this

• # highmarker

Posted September 21, 2012 - 07:10 AM

fly by wire throttles can be used to reduce compression braking by shutting the fuel off but leaving the throttle plate open for a time. Used in 1000cc 4 stroke yamaha snowmobiles so you don't go over the bars when you let off, or lock the track up on ice.

corner to corner the rekluse pretty much acts the same as any bike IMO, crawling down a long steep loose hill is where they will free wheel (and over heat brakes) .

• # grayracer513

Posted September 21, 2012 - 08:31 AM

So if I understand you correctly, you are saying that after the string breaks the ball will continue to turn and travel in a direction other than the V vector at the point of the string breaking? What is applying this force to make it turn?

Yes. The force that causes it is the inertia built up within the mass as a reaction to the lateral acceleration imposed by the string. That's stored energy that has to go somewhere in the sudden absence of the resistance of the string.

• # grayracer513

Posted September 21, 2012 - 08:33 AM

I'm not really sure how you can dispute this

Try to duplicate it in any practical way. I actually tried to make the weight fly to the left when released at 12:00, but it always goes to 12:00. According to that post, It should go to 12:00 if released at 3:00, but it goes to 3:00.

fly by wire throttles can be used to reduce compression braking by shutting the fuel off but leaving the throttle plate open for a time. Used in 1000cc 4 stroke yamaha snowmobiles so you don't go over the bars when you let off, or lock the track up on ice.

AFIK, that's used on the current big YZF-R's in place of a BTL clutch. The "dashpot" effect. Much better for the drivetrain, more effective, and programmable. Obviously no help to the OP, though.

• # KJ790

Posted September 21, 2012 - 08:45 AM

Yes. The force that causes it is the inertia built up within the mass as a reaction to the lateral acceleration imposed by the string. That's stored energy that has to go somewhere in the sudden absence of the resistance of the string.

But inertia equals mass times velocity, so inertia always acts in the same direction as the velocity vector in the diagram above. Inertia will not make it change direction after the string breaks.

Try to duplicate it in any practical way. I actually tried to make the weight fly to the left when released at 12:00, but it always goes to 12:00. According to that post, It should go to 12:00 if released at 3:00, but it goes to 3:00.

You lost me again. When the string breaks the ball would not fly towards any point of the clock, it would fly away from the clock on a path tangent to the original circular motion and perpendicular to the string's position at the breaking point. As soon as the string breaks the ball goes into straight line motion. As far as the ball is concerned it was like there never was any circular motion involved.

I can't find a single diagram, deomonstration, or equation anywhere suggesting otherwise. These are the laws used to put men on the moon. They are not teaching fake physics in schools.

Build a catapult with a cup on the end of the arm to hold a ball. Make it spring loaded so it will swing from 6:00 to 3:00 and then hit a stop. When the arm stops at 3:00 (horizontal) the ball will come out of the cup, go straight into the air, and come straight back down into the cup. It does not matter that the ball was in circular motion before the arm came to a stop. The arm hitting the stop is exactly the same as the string breaking in your previous example.

• # grayracer513

Posted September 21, 2012 - 09:30 AM

But inertia equals mass times velocity, so inertia always acts in the same direction as the velocity vector in the diagram above. Inertia will not make it change direction after the string breaks.

You lost me again. When the string breaks the ball would not fly towards any point of the clock, it would fly away from the clock on a path tangent to the original circular motion and perpendicular to the string's position at the breaking point. As soon as the string breaks the ball goes into straight line motion. As far as the ball is concerned it was like there never was any circular motion involved.

I can't find a single diagram, deomonstration, or equation anywhere suggesting otherwise. These are the laws used to put men on the moon. They are not teaching fake physics in schools.

Build a catapult with a cup on the end of the arm to hold a ball. Make it spring loaded so it will swing from 6:00 to 3:00 and then hit a stop. When the arm stops at 3:00 (horizontal) the ball will come out of the cup, go straight into the air, and come straight back down into the cup. It does not matter that the ball was in circular motion before the arm came to a stop. The arm hitting the stop is exactly the same as the string breaking in your previous example.

OK, whatever. When you (or anyone) can make that happen with the weight on a string, you let me know.

• # KJ790

Posted September 21, 2012 - 09:56 AM

OK, whatever. When you (or anyone) can make that happen with the weight on a string, you let me know.

Like I said, you can find demonstrations showing exactly what happens. You can see it everytime a baseball is hit, everytime a catapult launches, and everytime a ball is thrown. I can't find a single thing that suggests what you are trying to say, demonstration or theory. If what you say is true, where are all of the demonstrations depicting it? Where are all of the other sources mentioning this phenomena that physics cannot explain?

The theory is backed up by every demonstration I have ever seen. There are not any invisible forces at work that are not backed up by Newton's laws. Draw a free-body diagram of the ball at any point in the example and you will see that centrifugal force is at no time acting upon the ball.

Edited by KJ790, September 21, 2012 - 10:31 AM.

• # bg10459

Posted September 21, 2012 - 01:39 PM

Try to duplicate it in any practical way. I actually tried to make the weight fly to the left when released at 12:00, but it always goes to 12:00. According to that post, It should go to 12:00 if released at 3:00, but it goes to 3:00.

It's been a REALLY long time since I've used any of this, so forgive me if my terminology is incorrect. I also searched for an easy to see demo, but, of course, nothing is easy.

First, I hope we can agree that once the ball is released it does not move along a curved path, other than the curve created by gravity. That is shown in this illustration:

Now maybe this is where Gray is going with his argument. If there is an acceleration (or deceleration) at the point of release, the ball will move at an angle other than perpendicular to the string, as shown in this illustration, but still in a straight line:

Or maybe it's the perception of the thrower that the ball moves along a curved path, eg. the coriolis effect as shown in this video

Edited by bg10459, September 22, 2012 - 01:20 PM.

• # MrN2OBelvedere

Posted October 12, 2012 - 06:38 PM

Actually, the ball could be sitting absolutely still (WRT a global coord. system) and everything else (the rest of the universe) spinning, rotating, accelerating around the ball. There'd be no difference really.

Anyone who's really interested in this on this forum (and I really think there are very few, considering most dirtbikers can't even talk about the attributes of a camshaft, much less kinematics and dynamics) could easily just google it. Go to wikipedia and look up angular velocity/acceleration and you will find all you need to know.

As for the engine braking bit, does this thing have higher compression or a lighter rotating assy than your 2006? I thought the 2006-09 were pretty similar.

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