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jcw
Starting Member
7 Posts
Chicago, IL
USA
BMW
K12R
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 Posted - 09/08/2009 : 11:44 PM
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Hello.
I recently had an off track excursion at a trackday after locking my front brakes on a BMW k1200r. I feel it was due to an overly quick application of the front brakes after a very quick acceleration up to the corner. The thing is, I'm sure I've braked as hard on a conventional front suspension bike without lock up previously.
I had some theoretical questions regarding BMW's choice of suspension/bike design and braking. The BMW has a duolever front suspension with "advertised" antidive. The wheelbase is also rather long and again they "advertise" a very low center of gravity.
1) I understand that antidive increases weight transfer by maintaining or raising the center of gravity. However, does the speed that this weight transfer occurs due to the lack of suspension travel make it more prone to lock up? Would a conventional "pro-dive" suspension tolerated my overly quick application of brakes better?
2) Does weight transfer increase or decrease overall braking performance? Does a tall/short wheelbase race bike brake better than my long/low BMW? It would seems that weight transfer balances out front to rear; however, I seem to recall that this may not be the case.
Look forward to your expert answers. 
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greywolf
 Advanced Member
686 Posts
[Mentor]
Evanston, IL
USA
Suzuki
DL650AK7
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Posted - 09/09/2009 : 12:51 AM
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| Weight transfer happens whether the front compresses or not. There was a BMW superbike racer in the late 1970s that had an anti dive front brake. It was competitive. You may have braked harder before, and you may not have. The only thing you know for sure was your braking force exceeded the available traction at that time in that place. I fitted an anti dive front brake mechanism on a CX500 Honda thirty years ago. It's braking ability was unchanged even after the dive was gone. You mentioned an overly quick application. That can be the main problem. The front brake application needs to be progressive. A small but finite amount of time needs to pass for the weight transfer to take place before maximum braking force can be applied. |
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jcw
Starting Member
7 Posts
Chicago, IL
USA
BMW
K12R
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Posted - 09/09/2009 : 7:17 AM
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I agree that the application of the brake too quickly is my problem.
But my thoughts on antidive are the following...
the braking force and weight transfer is initially shared by the compression of the forks and the tires in a pro-dive suspension easing the load on the tire.
Or an antidive, the total load is immediately borne by the tires.
Ultimately, both will bear about the same weight (if CoG is the same) but I wonder if the initial application needs to be extra easy on an anti dive suspension? |
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gymnast
Moderator
2814 Posts
[Mentor]
Meridian, Idaho
USA
Harley-Davidson
Sportster Sport
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Posted - 09/09/2009 : 8:24 AM
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As greywolf points out, anti-dive forks were thought to be the "hot set up" in the 1970s and all the major manufacturers seemed to offer some version of the theme on one or another of their bikes including Harley Davidson. The bike magazines touted anti dive forks as a must have for improved performance. Just as quickly it seemed all mention of anti dive forks stopped and the emphasis became custom fork valving set up's. Both trends seem to have been driven by an attempt to please the advertisers by creating a need that few actually had for the touted products.
It appears that you may need to learn to adjust and adapt your braking technique through gradual increments when getting on a machine that is different than your everyday rider as implied by your comments above. Taking your time and adapting to your equipment is, perhaps, more germane to preventing lock up than sorting out theoretical as well as questions of applied physics while "running hot into a corner.
Did your track day instructors give you any guidance after you ran off the track?
Both of your questions are so generally framed so as to provide more for a general discussion than specific answers. In some cases you seem to be throwing in apples with the oranges.
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jcw
Starting Member
7 Posts
Chicago, IL
USA
BMW
K12R
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Posted - 09/09/2009 : 10:20 AM
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I really wasn't running too hot into the corner. I was going fast but composed (for the track). It kinda caught me off guard, really. I felt the application of the brake SHOULD have been well tolerated by the suspension(but maybe not this suspension design, though, which is what I'm trying to wrap my mind around). Granted, it wasn't the best application of braking, but I thought it was well within the limits of the machine.
The BMW is my "daily" rider but on the streets I simply do not apply the brakes as hard or as quickly as I might do on the track. And it's hard to practice threshold braking without risking bike, life and limb on the streets.
And that may be my issue.
I agree that attention to fork valving has replaced antidive technology in the majority of motorcycles (except BMW's). The fact that antidive is not on any current MotoGP bikes kinda suggests that they are not needed but who knows if it is driven by marketing and money.
I hope I'm not coming off too stand-off-ish. I just wanted to maybe hear some scientific explanations of this process. I've been kicked off of more than one board for asking the "wrong" questions before. It's not my intention to piss anyone off.  |
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James R. Davis
Administrator
14935 Posts
[Mentor]
Houston, TX
USA
Honda
GoldWing 1500
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Posted - 09/09/2009 : 10:31 AM
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quote:
But my thoughts on antidive are the following...
the braking force and weight transfer is initially shared by the compression of the forks and the tires in a pro-dive suspension easing the load on the tire.
Well, let's see if this helps. Whether 'weight transfer is initially shared by compression of the forks and tires', as you say, the fact remains that when weight is transferred from the rear tire it is THEN added to the front tire. That weight is not 'lost' or somehow swallowed by the springs in your shocks. The combined weight on the front and rear tires remains the total weight of the bike and rider at all times unless the roadway falls away from the bike or rushes up to meet that bike.
And, with anti-dive, the amount of weight that is transferred for any given rate of deceleration is higher than on a non-anti-dive bike because the CG remains higher during braking while the wheelbase remains the same. |
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galileo
Ex-Member
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Posted - 09/09/2009 : 11:35 AM
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quote:
Originally posted by James R. Davis
quote:
But my thoughts on antidive are the following...
the braking force and weight transfer is initially shared by the compression of the forks and the tires in a pro-dive suspension easing the load on the tire.
Well, let's see if this helps. Whether 'weight transfer is initially shared by compression of the forks and tires', as you say, the fact remains that when weight is transferred from the rear tire it is THEN added to the front tire. That weight is not 'lost' or somehow swallowed by the springs in your shocks. The combined weight on the front and rear tires remains the total weight of the bike and rider at all times unless the roadway falls away from the bike or rushes up to meet that bike.
And, with anti-dive, the amount of weight that is transferred for any given rate of deceleration is higher than on a non-anti-dive bike because the CG remains higher during braking while the wheelbase remains the same.
After an hour and a half of typing and deleting, I have a hypothesis. The load transferred is always a function of the horizontal distance from the front wheel to the CG divided by the height of the CG. (P. 99 of Motorcycle Dynamics.) You say the CG is low. With the location of the engine on that bike, I suspect it is also a little forward, although this doesn't have to be so. How these actually balance would have to be measured.
Putting this in english, the percentage load on the front tires during maximum braking is affected by the geometry of the bike. You would have to know the location of the CG to determine the effect.
It may be that your previous bike had a higher percentage of the load on the front brake during braking. Say 90%. The amount of deceleration that can be obtained using only the front brake is the percentage load on the front wheel times the coefficient of friction. Assuming a CF of 1, that would be .9 in this case. Any attempt to brake harder to this would result in the front wheel locknig.
With a low, forward CG it may be that only 75% of the weight is transferred to the front wheel during braking and the front wheel would lock at .75g's if only the front brake is used. (75% used only as an example.)
If you were judging how hard your braking was based on feel, or always braking at a certain speed at a certain distance from the curve, it's quite possible this was the cause of your lockup.
I'd talk to others with a bike similar to yours to find out how they do it. I don't want to make suggestions as the track is much more serious than a day in the parking lot. There are many things I don't know about the track. Like most of them.
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James R. Davis
Administrator
14935 Posts
[Mentor]
Houston, TX
USA
Honda
GoldWing 1500
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Posted - 09/09/2009 : 12:18 PM
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quote:
The load transferred is always a function of the horizontal distance from the front wheel to the CG divided by the height of the CG.
What in the world are you talking about?. That is simply WRONG.
What you apparently mean is how prone your bike is to doing an end-over is largely a function of the front or rear bias of the CG.
Weight transfer is a function of how HIGH the CG is as compared to the length of the wheelbase. Where the CG is located (front or rear bias) is an improper focus in this discussion.
A given rate of deceleration depends only on the ratio of the height of the CG to the length of the wheelbase to determine what percentage of the total weight of the bike and rider is moved forward. The forward or rear bias of the CG only determines what the starting (at rest) distribution of that total weight is on each tire.
So, for example, if 60% of the weight of the bike is on the rear tire (rear bias 40/60) and the total weight is 1000 pounds, then at rest 600 pounds is on the rear tire and 400 pounds is on the front tire. Let's say that the ratio of Height of CG to length of wheelbase is .5 and you achieve a deceleration rate of 0.9g's. That means that 50% 45% of the total weight of the bike (in other words, 500 450 pounds) will be shifted from the front tire from the rear. In our case, the bike does not end-over. But if the weight bias was 60/40 at rest, then there would have been only 400 pounds on the rear tire and removing 500 pounds would lift that wheel off the ground with an end-over in progress. In either case the deceleration rate was 0.9 and the amount of weight transfer was the same.
What your focus on front or rear bias does is determine how prone your bike is to doing an end-over - not 'the load transferred'.
I am even more mystified by your joining the concepts of forward bias of the CG with a LOW CG and IMPLYING that they have similar effects. A low CG diminishes loading or weight transfer making it difficult or impossible to do an end-over. A forward CG bias increases the ability to do an end-over.
Edited to correct the math - JRD |
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jcw
Starting Member
7 Posts
Chicago, IL
USA
BMW
K12R
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Posted - 09/09/2009 : 1:20 PM
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I understand what he meant- load transfer is a function of height of CofG divided by horizontal distance of tire contact patch to CofG. (not the other way around)
In other words, load or weight transfer is increased in bikes with a higher CofG AND in bikes with a shorter wheelbase.
I guess the horizontal position of the C of G is important only in the static weights on the front and rear wheel. But doesn't contribute to the weight transfer.
Part of my original question though was is weight transfer to the front wheel desirable to overall braking force. Or is something loss with that transfer. |
Edited by - jcw on 09/09/2009 1:31 PM |
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galileo
Ex-Member
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Posted - 09/09/2009 : 1:23 PM
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James,
I agree with everything you say except that my focus was on doing an end over. Although based on the page reference, I can see how you got that opinion.
I was not inferring that a forward CG reduces the probability of an end over. It doesn't. I was inferring that his CG appears forward, but it may not be. I should have left the following statement out as I agree it was confusing.
quote:
With the location of the engine on that bike, I suspect it is also a little forward, although this doesn't have to be so. How these actually balance would have to be measured.
I should have referenced the equations on page 101. For optimum braking the relationships are as follows. Attempting to exceed this optimum by either tire will result in a skid on that tire.
Ff/F = (b+uh)/p
Fr/F= ((p-b)-uh)/p
What I'm suggesting is the value for Ff/F was higher and the value for Fr/F was lower for optimum braking on his previous bike. Not realizing this, he may have locked the front brake in an effort to get too much force out of the front brake.
It becomes more complicated if he had already entered the turn and that is beyond my current knowledge. It also ignores aerodynamic forces which would change the numbers, but I don't think it would change the concept.
quote:
Let's say that the ratio of Height of CG to length of wheelbase is .5 and you achieve a deceleration rate of 0.9g's. That means that 50% of the total weight of the bike (in other words, 500 pounds) will be shifted from the front tire from the rear.
I think you may have meant to say 45% or 450 pounds. I think 500 pounds would occur at 1 g.
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James R. Davis
Administrator
14935 Posts
[Mentor]
Houston, TX
USA
Honda
GoldWing 1500
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Posted - 09/09/2009 : 1:30 PM
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quote:
I think you may have meant to say 45% or 450 pounds. I think 500 pounds would occur at 1 g.
Yep - if the rate of deceleration was 0.9 then the transfer would be 450 as you said. Typed too quickly. Mea culpa. |
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jcw
Starting Member
7 Posts
Chicago, IL
USA
BMW
K12R
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Posted - 09/09/2009 : 1:39 PM
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I agree totally with this. And I understand the concept. From my post above...
Ultimately, both will bear about the same weight (if CoG is the same) but I wonder if the initial application needs to be extra easy on an anti dive suspension?
So both when a "steady state" is reached should be loaded a similar amount with the anti dive loaded slightly more due to the higher CofG and longer wheelbase.
But, my lockup was in that instant the front brake was pulled. It seems to me that on an antidive susupension design, the lack of front suspension movement to gradually load the front tire makes it more prone to lock up than a conventional fork without antidive.
I just need to make a more softer intial brake movement.
quote:
Originally posted by James R. Davis
quote:
But my thoughts on antidive are the following...
the braking force and weight transfer is initially shared by the compression of the forks and the tires in a pro-dive suspension easing the load on the tire.
Well, let's see if this helps. Whether 'weight transfer is initially shared by compression of the forks and tires', as you say, the fact remains that when weight is transferred from the rear tire it is THEN added to the front tire. That weight is not 'lost' or somehow swallowed by the springs in your shocks. The combined weight on the front and rear tires remains the total weight of the bike and rider at all times unless the roadway falls away from the bike or rushes up to meet that bike.
And, with anti-dive, the amount of weight that is transferred for any given rate of deceleration is higher than on a non-anti-dive bike because the CG remains higher during braking while the wheelbase remains the same.
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James R. Davis
Administrator
14935 Posts
[Mentor]
Houston, TX
USA
Honda
GoldWing 1500
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Posted - 09/09/2009 : 1:43 PM
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quote:
Part of my original question though was is weight transfer to the front wheel desirable to overall braking force.
Since the amount of traction you have is a function of weight carried by the tire, obviously weight transfer to the front tire is RADICALLY important (and desireable) to overall braking. |
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jcw
Starting Member
7 Posts
Chicago, IL
USA
BMW
K12R
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Posted - 09/09/2009 : 1:43 PM
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quote:
Originally posted by gymnast
Did your track day instructors give you any guidance after you ran off the track?
You grabbed too much front brake.  |
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jcw
Starting Member
7 Posts
Chicago, IL
USA
BMW
K12R
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Posted - 09/09/2009 : 1:52 PM
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quote:
Originally posted by James R. Davis
quote:
Part of my original question though was is weight transfer to the front wheel desirable to overall braking force.
Since the amount of traction you have is a function of weight carried by the tire, obviously weight transfer to the front tire is RADICALLY important (and desireable) to overall braking.
This is where I'm confused. Lee Parks in his book on motorcycle riding states that the goldwings, harleys and other long wheelbase bikes could easily outbrake a sportbike up to the recent past.
These bike have less weight transfer due to the longer wheelbase.
From the book- motorcycle design and dynamics (?) I also understand that perhaps threshold braking is limited by one of two things...
1) you've reached the limit in front tire adhesion OR
2) you start to lift the rear wheel at which time if you don't lift off the brake a little you will continue over to do an endo.
What was my question, in the first place? |
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James R. Davis
Administrator
14935 Posts
[Mentor]
Houston, TX
USA
Honda
GoldWing 1500
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Posted - 09/09/2009 : 2:38 PM
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quote:
This is where I'm confused. Lee Parks in his book on motorcycle riding states that the goldwings, harleys and other long wheelbase bikes could easily outbrake a sportbike up to the recent past.
These bike have less weight transfer due to the longer wheelbase.
Lee Parks has great credentials on the race track, but his discussion of motorcycle dynamics leaves a lot to be desired.
It is true that sport bikes TEND not to be able to out brake other styles of bikes, it is not because these others have 'less weight transfer'. I assure you that a Goldwing will invariably transfer more weight at a given deceleration rate than ANY sport bike, for example.
The reason is that the ratio of the height of the CG as compared to length of wheelbase of a sport bike is higher than for these other bikes. What actually limits braking performance of bikes is the amount of traction they can utilize while braking. As soon as a sport bike lifts its rear wheel off the ground it cannot achieve any higher rate of deceleration - despite having still unused front tire more traction. Having that higher ratio means that a sport bike transfers a greater percentage of its total weight from the rear to the front than the others at a given rate of deceleration. That means that a sport bike can do a stoppie more easily than the others and thus limit its own braking ability.
It is also not true that only sport bikes can do a stoppie. Though it is harder for the others to do so, many of them can. If and when they do, however, they have already achieved a higher rate of deceleration than a typical sport bike.
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Andrew Dressel
Standard Member
153 Posts
Milwaukee, WI
USA
Moto Guzzi
California Special
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Posted - 09/10/2009 : 12:19 PM
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quote:
Originally posted by James R. Davis
Weight transfer is a function of how HIGH the CG is as compared to the length of the wheelbase.
In attempt to tease out exactly what role CM location (vertical and horizontal) has on braking performance, I've attempted to create graphs that show how braking force (assuming a coefficient of friction of 1.0 and perfect threshold braking) varies as the CM moves along both axes. I'm using Cossalter's convention of "h" for CM height above the ground and "b" for CM distance forward of the rear hub/contact-patch-center. I'm also using his numbers from page 99-101 of 200 kg mass and 1.35 meter wheelbase.
I do not display the surfaces wherever conditions would produce tipping: rear wheel normal force goes to zero.
As always, I look forward to comments and will do my best to correct errors as soon as they are discovered. |
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James R. Davis
Administrator
14935 Posts
[Mentor]
Houston, TX
USA
Honda
GoldWing 1500
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Posted - 09/10/2009 : 12:37 PM
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I have real trouble with people posting charts that are not explained, do not have proper legends, are derived from formulas that are not shown, and are posted following a quote as if they either support or challenge that quote, but do neither.
Both charts, for example, claim to show something in red. There is no red shown.
What are our readers to learn from these charts?
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Andrew Dressel
Standard Member
153 Posts
Milwaukee, WI
USA
Moto Guzzi
California Special
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Posted - 09/10/2009 : 1:30 PM
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quote:
Originally posted by James R. Davis
I have real trouble with people posting charts that are not explained, do not have proper legends, are derived from formulas that are not shown, and are posted following a quote as if they either support or challenge that quote, but do neither.
Both charts, for example, claim to show something in red. There is no red shown.
What are our readers to learn from these charts?
Ouch, sorry James.
1. I thought I did explain the charts. I told how "h" and "b" are defined. I told what mass, wheelbase, and coefficient of friction I used. I told what the three different color surface represent.
2. Each has labels on all three axes with units. What is missing or improper?
3. The one on the right is derived from the formulas provided by Cossalter for the normal force generated when both wheels brake at threshold and listed in a posting above by galileo. The other is from expressions derived the same way, but for front and rear wheel braking separately.
4. I posted them in response to the discussion of the role played by CM location. I find it helpful to see a picture of something like that, I couldn't find one, so I made one. I thought others might like to see it too. In the past I have received positive feedback for similar efforts.
5. On my monitor, the top triangle is red in both pictures.
6. I cannot say for sure what others might learn or not, but what I learned from the exercise, though I certainly should already know it, is how important braking with both wheels is for best results short of performing a stoppie.
Added: One problem could be how the image has been rendered after I posted it. In the preview, I saw it full-sized, and the red is much easier to see. As I see it rendered now, it is hard to tell "h" from "b" on the axes. The original image is available at this url: https://pantherfile.uwm.edu/adresse...analysis.png
If there is some way that I can format the image so that it is rendered better here, please let me know. |
Edited by - Andrew Dressel on 09/10/2009 9:05 PM |
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galileo
Ex-Member
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Posted - 09/10/2009 : 1:32 PM
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Andrew,
This might be better placed in the Motercycle Braking Dynamics thread.
I'm really interested that you gave me a new math puzzle. If I were doing this, I'd present it a little differently. Consider this brainstorming and there may be an error in my logic.
I'd try something with the equations on page 101. Rather than using actual dimensions, cancel out a few things.
For 1g braking, F will equal weight. Just make it 1. Then Ff will be a percentage of the weight. As will Fr later. Ff and Fr will be a percentage of total braking.
Express b and h as a percentage of p which will eliminate p from the equation.
So you are left with Ff=B+h where Ff is a percentage of total braking.
Make a chart with b/p on the bottom and h/p on the vertical. Make a scattergram chart. Then one could drag the point around and see the effect. You would have to use VBA to make a label for the point giving the percentage of braking for each tire. I can do this, but not now as I'm off for a ride shortly.
Added: I have an idea that I could superimpose a picture of a bike as a background on the chart and then one could use click arrows to move the CG. The percentage braking for each wheel would show under each wheel. The reason this would be useful is it could help show any change by adding a passenger or loading the bike. It might also be applicable in determining changes in handling due to shifting the CG. But I'm not into that area yet.
Neat project. I may get to it tonight. |
Edited by - galileo on 09/10/2009 1:43 PM |
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James R. Davis
Administrator
14935 Posts
[Mentor]
Houston, TX
USA
Honda
GoldWing 1500
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Posted - 09/10/2009 : 9:14 PM
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quote:
If there is some way that I can format the image so that it is rendered better here, please let me know.
The image is a png file which, although I obviously can handle it, is not really consistent with the design of my software (which was intended to handle jpg and gif formats). Nevertheless, I went to the original diagrams and can see that the 'red' actually consists of half the pixels being a dark red while the other half being black. The result is a virtually black rendering.
As to positive feedback ... you get that for your chart construction skills, but I suggest that of the 200+ readers here who have looked at it there might be five who can figure out what it tells them or how to use it.
The quote that preceded it was about weight transfer being a function of height of CG as compared to length of wheelbase. Does either chart show that or dispute that? If either, shouldn't you point that out, else why use the quote?
Instead, you say:
quote:
6. I cannot say for sure what others might learn or not, but what I learned from the exercise, though I certainly should already know it, is how important braking with both wheels is for best results short of performing a stoppie.
How hard would it be to say something like, "looking at the chart on the left you can see that ... and where the x-axis crosses the y-axis there is evidence that ..." so that our non physics or non math majors can hope to 'get it'? I am no slouch when it comes to physics and math and I am pretty handy at reading charts, but I do not 'get it' from what you posted. Maybe it's just me.
In any event, I really dislike being made to feel like a dummy and I suspect the same is true of our other members. |
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Question about weight transfer, anti-dive and braking