(Please visit one of our advertisers)

No donations or subscriptions are required

   OR   
   
Subscription choices:
Board Karma = 40  (3488 positive of 3870 votes is 40 %pts higher than a neutral 50%)
All Things (Safety Oriented) Motorcycle   
Username:
Password:
Save Password
Forgot your Password?

You can the entire collection of Safety Tip articles in a 33 Megabyte PDF Portfolio

 All Forums
 Motorcycle Safety
 Physics and the theoretical
 Turning, center of gravity and camber thrust
Previous Page | Next Page
Member Previous Topic Discussion Topic Next Topic
Page: of 3

gymnast
Moderator
4265 Posts
[Mentor]


Meridian, Idaho
USA

Harley-Davidson

Sportster Sport

Posted - 07/25/2013 :  9:40 AM
The effect of camber thrust becomes more of a factor if the rear wheel is out of alignment in it's mounts or the frame is bent. Another factor which may contribute to noticeable camber thrust problems is worn swing arm bearings.
Go to Top of Page

James R. Davis
Male Administrator
17286 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 07/26/2013 :  4:05 PM Follow poster on Twitter  Join poster on Facebook as Friend  
quote:
Originally posted by The Meromorph

It's a relatively minor point, but the rear tire does exert a turning force under camber thrust. No-one is claiming that the rear wheel turns relative to the frame, but the frame turns (to some extent 'dragged round' by the steering head) and the camber thrust additively assists in this process.
The greater the contribution of camber thrust from the rear tire, the 'easier' the steering feels. Changing the rear tire from a square to a round profile causes the steering to feel 'quicker'.




Apples and oranges.

Because the rear tire is completely constrained from rotating relative to the frame, camber thrust CANNOT aid or detract as it does not exist. On the other hand, if the rear tire profile was flat as opposed to rounded, turning would feel less smooth than if the tire profile were rounded, but that is because a turn requires the rear-end of the bike to be raised during that turn as the contact patch becomes the edge of the flat surface as seen below.

Go to Top of Page

Andrew Dressel
Male Standard Member
244 Posts


Milwaukee, WI
USA

Moto Guzzi

California Special

Posted - 07/26/2013 :  5:29 PM
quote:
Originally posted by James R. Davis
Because the rear tire is completely constrained from rotating relative to the frame, camber thrust CANNOT aid or detract as it does not exist.


Sorry for coming late to the party, and sorry if I misunderstand Jame's statement above, but I don't see how this constraint prevents camber thrust from occurring at the rear tire.

All that is required for camber thrust to occur is for the tire to roll forward at a non-zero camber angle. That is how it is measured in the laboratory.

In addition to Cossalter's test device, pointed out by JanK above, which has recently been used on bicycle tires (A. Doria, et al., "Identification of the mechanical properties of bicycle tyres for modelling of bicycle dynamics," Vehicle System Dynamics, 2012), I also happen to have constructed several devices for measuring bicycle tire camber thrust, both here in Milwaukee and in the Netherlands. I've posted some pictures on flikr at http://www.flickr.com/photos/andrew...34618198727/

Even narrow bicycle tires, just 22 mm wide, generate camber thrust just fine.

The main reason I have been working so hard for the past couple of years to measure camber thrust and corning force separately is that each produces a different torque about a vertical axis through the contact patch, as Cossalter explains in Motorcycle Dynamics. This doesn't matter too much at the rear wheel, because of the constraint that James points out above, but we expect that it can have a large influence over the torque required at the handlebars from the rider to maintain a steady-state turn.

A big question is whether camber thrust is sufficient to produce all the side force necessary or is deficient and must be supplemented by corning force from a non-zero slip angle. Cossalter actually finds it to be slightly excessive on motorcycle tires so that they have a slightly negative slip angle to maintain equilibrium. Doria finds a similar results on 3 out of 4 bicycle tires. I have found it so far to be usually deficient on bicycle tires. I have not yet been able to explain the discrepancy between my results and Doria's.

I am hoping to meet up with someone from the Padua team at the upcoming Bicycle and Motorcycle Dynamics Conference this November at Nihon University in Japan. Perhaps we can sort it out then.

Except for the different torques about a vertical axis through the contact patch generated by camber thrust and corning force, I don't see how it matters much to the bike which is responsible for the side force generated by the tires. Numerical models based on knife-edge wheels that roll without slip, and so ignore both camber thrust and corning force, accurately predict, as validated by physical testing, the capsize and weave modes of an uncontrolled bike.

Edited by - Andrew Dressel on 07/28/2013 7:57 PM
Go to Top of Page

scottrnelson
Advanced Member
6888 Posts
[Mentor]


Pleasanton, CA
USA

KTM

990 Adv, XR650L

Posted - 07/26/2013 :  6:21 PM
quote:
Originally posted by Andrew Dressel

...I also happen to have constructed several devices for measuring bicycle tire camber thrust...

Even narrow bicycle tires, just 22 mm wide, generate camber thrust just fine.
How much force is involved?

I haven't seen anybody post a number of pounds of force exerted by camber thrust - or I missed it, if it was buried deep in a bunch of other text.

So how many pounds of force can you get from a bicycle tire leaned over at 30 and 45 degrees? What about motorcycle tires?
Go to Top of Page

JanK
Male Junior Member
76 Posts


Ljubljana, Ljubljana
Slovenia

BMW

F650CS

Posted - 07/27/2013 :  6:11 AM
They are on the order of 1g. In the first example from Cossalter's text, where the camber force at 28o is sufficient to generate all the centripetal force (i.e., no slip), the camber acceleration would be g*tan(28o) = 0.53g.
Go to Top of Page

scottrnelson
Advanced Member
6888 Posts
[Mentor]


Pleasanton, CA
USA

KTM

990 Adv, XR650L

Posted - 07/27/2013 :  7:34 AM
quote:
Originally posted by JanK

They are on the order of 1g. In the first example from Cossalter's text, where the camber force at 28o is sufficient to generate all the centripetal force (i.e., no slip), the camber acceleration would be g*tan(28o) = 0.53g.

Sorry, but I can't believe those numbers. Camber thrust would be providing ALL of the turning force if that were the case.
Go to Top of Page

Andrew Dressel
Male Standard Member
244 Posts


Milwaukee, WI
USA

Moto Guzzi

California Special

Posted - 07/27/2013 :  9:52 AM
quote:
Originally posted by scottrnelson

quote:
Originally posted by JanK

They are on the order of 1g. In the first example from Cossalter's text, where the camber force at 28o is sufficient to generate all the centripetal force (i.e., no slip), the camber acceleration would be g*tan(28o) = 0.53g.

Sorry, but I can't believe those numbers. Camber thrust would be providing ALL of the turning force if that were the case.



Well, that's what Cossalter has reported from his test device, and even higher for 3 of the 4 bicycle tires Doria tested. In my testing, I have a couple tires, out of 14, that came out that high, but most were lower than than, but as much as half.

Another way the results are reported is in a normalized coefficient with units of 1/degree. To calculate the force generated, multiply the coefficient by the vertical load and by the camber or slip angle, as the case may be. I found cornering stiffness coefficients for bicycle tires to be around 0.15 to 0.3 and camber stiffness coefficients to be around 0.008 to 0.02 (both in 1/degree). The so-called "tangent rule", in which the camber thrust is exactly enough to provide all the lateral force required for equilibrium in a turn, is 0.017455 1/degree.

In any case, why the objection to camber thrust providing all of the lateral force?
Go to Top of Page

scottrnelson
Advanced Member
6888 Posts
[Mentor]


Pleasanton, CA
USA

KTM

990 Adv, XR650L

Posted - 07/27/2013 :  11:48 AM
quote:
Originally posted by Andrew Dressel

In any case, why the objection to camber thrust providing all of the lateral force?
Not really an objection, it's just something that I don't really understand, but I want to understand.

Maybe we need a clear definition of what camber thrust is, in case I'm thinking it's something else.

I've been led to believe that camber thrust was the turning behavior due to the conical shape of the tire when leaned over. The numbers in the most recent posts above have to be talking about something else. I have a hard time believing that a significant amount of the turning force is due to the part of the tire closer to the center being larger circumference than the part of the tire closer to the edge, and that the difference is what does most of the work while turning.

I've always felt that the turning force has more to do with how the front wheel is pointed and that the side forces on the tire are not significantly different from the side forces on a car tire when cornering, which most definitely don't do any of their turning due to some conical shape of the tire.

So somebody please provide a more concise definition of what camber thrust really is and I'll be happy to have you point out where my thinking is wrong or what it is that I'm misunderstanding.

All I really want is to understand what's really going on, and I'm always happy to change my opinions when presented with facts that contradict them.

I'm just after the truth.
Go to Top of Page

twowheelsbg
Male Junior Member
50 Posts


Burgas, Burgas
Bulgaria

Suzuki

Posted - 07/27/2013 :  1:16 PM
Try this link, Scott,
http://en.wikipedia.org/wiki/Camber_thrust

It is simply explained how leaned tire is pushed inside the corner,
that is lean causes cornering. Do not mix this effect with cornering coffee cups analogy,
it also exists, but is another effect -
the appied camber thrust point is slightly off the contact patch center.

Thanks for the article to Mr.Andrew Dressel
Go to Top of Page

James R. Davis
Male Administrator
17286 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 07/27/2013 :  2:44 PM Follow poster on Twitter  Join poster on Facebook as Friend  
Again, apples and oranges.

There are TWO phenomena that are both referred to as 'camber thrust'. In the simplest case, the one which I have described elsewhere here on the site and which was just described as the 'cornering coffee cups analogy', a cambered profile that is leaned results in the circumference of the tire closest to its mid-line traveling faster than the circumference of the tire away from the mid-line. That results in a MODEST torque of the wheel as it rolls which makes the tire try to point toward the inside of its turn. That force is really insignificant overall, especially since it is overwhelmed by the forces caused by deformation of the tire.

In other words, if the tire were not made of rubber or any other deformable material - if it was HARD - this form of 'camber thrust' would be meaningful.

Before we get into the other form of 'camber thrust', we need to recall that a rolling tire CANNOT change its direction of travel unless it can SLIP (and deform) with the surface it is rolling on. A tire that cannot be deformed and that cannot slip relative to the roadway surface MUST roll in a straight line.

In a turn, the difference between the direction the wheel is pointing and its actual direction of travel is called its 'slip angle'.

Note the absolute dependence on tire deformity. And now a reminder, without deformity, only slip can be used to change the tire's path of travel - which is what the 'rolling coffee cup' demonstrates.

In addition to some part of the contact patch slipping on the surface more than other parts of that contact patch, lateral forces cause the contact patch to deform such that the 'outside' edge of the contact patch becomes compressed (and the whole contact patch moves toward the inside of the turn and out from under the mid-line of the wheel).

That deformity is slightly different than what was described above as 'tire slip'. It is what is called the 'camber angle' as opposed to the 'slip angle' and is the second source of 'camber thrust' that has been introduced into this discussion. Indeed, they are both 'camber thrust', but deformation of the tire exaggerates the 'rolling coffee cup' analogy such that one is hard pressed to explain it in terms that are accessible to the layman.

'Camber thrust' based on the consequences of the profile of the tire is what I have talked about here and elsewhere on the site. It does NOT depend on tire deformity. 'Camber thrust' as now being discussed not only depends on the profile of the tire, it depends on weight, lean angle, tire pressure, tire construction (biased vs radial), tire compounds, and road surface.

Keep it simple and more people will understand the concepts. Make it complex and readers go elsewhere. [edit: I eliminated/changed what originally sounded too much like a slam. JRD]

Go to Top of Page

Andrew Dressel
Male Standard Member
244 Posts


Milwaukee, WI
USA

Moto Guzzi

California Special

Posted - 07/28/2013 :  4:28 PM
quote:
Originally posted by James R. Davis

Again, apples and oranges.

There are TWO phenomena that are both referred to as 'camber thrust'.



Ah ha! Excellent. That helps a lot. I am familiar with both phenomena, and we've measured both in the lab, but I have only known "camber thrust" and "camber force" to mean the lateral force towards the direction of the lean. The torque about a vertical axis that tends to turn the wheel toward the direction of the lean is called "twisting torque" by Cossalter, and merely "Mz" by Pacejka. Even Foale, in his presentation of the cone model only uses camber force to describe force, as far as I can tell, not torque. I can't see if he even mentions the torque generated in the contact patch in his book.

Forgive me for misreading your Camber Thrust glossary entry. I took "attempt to make the bike turn a tighter radius" to indicate "centripetal force". I didn't get that you were referring to the torque generated in the contact patch. I certainly had no intention to mix apples and oranges to complicate things.

Okay. With that sorted out, now I totally get what you mean about the torque generated in the contact patch due to camber in the rear tire to not matter much at all because the rear tire is constrained by the rear frame.

As for how much the torque generated in the contact patch due to camber contributes to the steering torque, that is harder to say. The coefficients we measure for torque due to camber are about 2.5% of the coefficients we measure for torque due to slip and opposite in sign. If the force due to camber is exactly equal to the total centripetal force required to maintain equilibrium in a steady-state turn, then slip will be zero, and there will be no torque due to slip. If camber force is too low such that a slip angle is necessary for equilibrium, then the torque created by 37o of camber would be canceled by the torque created by just 1o of slip. If camber force is too high, requiring a negative slip angle as Cossalter supposes, then the torques would add at equally disproportionate rates. I'm afraid the real answer for any situation would depend on the tire and the bike.

Anyway, to answers Scott's question about magnitude, at least for bicycle tires, an example coefficient we recently measured is 0.0268 N-m/o. So, yeah, pretty small. Doria appears to report similar numbers for bicycle tires, as does Cossalter for motorcycle tires.

Finally, in answer one of Lilley's original questions about how bicycle tires can generate enough camber force even though they are so thin, a different model of the tire, due to Rotta, supposes that the lateral force can be calculated by multiplying the inflation pressure, which is much higher in skinny bicycle tires, by the difference in the radii of the side bulges on either side of the contact patch.

[edit: Improve word order, fix spelling, and fix formating. AED]

Edited by - Andrew Dressel on 07/28/2013 9:12 PM
Go to Top of Page

scottrnelson
Advanced Member
6888 Posts
[Mentor]


Pleasanton, CA
USA

KTM

990 Adv, XR650L

Posted - 07/28/2013 :  7:39 PM
quote:
Originally posted by Andrew Dressel

Anyway, to answers Scott's question about magnitude, at least for bicycle tires, an example coefficient we recently measured is 0.0268 N-m/o. So, yeah, pretty small. Doria appears to report similar numbers for bicycle tires, as does Cossalter for motorcycle tires.
Any chance you can convert that to some units commonly used in the USA, for 30 degrees and for 45 degrees?

Thanks.
Go to Top of Page

Andrew Dressel
Male Standard Member
244 Posts


Milwaukee, WI
USA

Moto Guzzi

California Special

Posted - 07/28/2013 :  9:04 PM
quote:
Originally posted by scottrnelson

quote:
Originally posted by Andrew Dressel

Anyway, to answers Scott's question about magnitude, at least for bicycle tires, an example coefficient we recently measured is 0.0268 N-m/o. So, yeah, pretty small. Doria appears to report similar numbers for bicycle tires, as does Cossalter for motorcycle tires.
Any chance you can convert that to some units commonly used in the USA, for 30 degrees and for 45 degrees?

Thanks.



Sure.

For a 35-mm-wide bicycle tire at 58 psi (4 bar) and under a 90 lb (41 kg) vertical load, Doria measured, on the 3-meter-diameter disk in Padua, a normalized cornering stiffness of 0.200 [1/o] and a self-aligning torque stiffness of 0.005 [m/o] at 1o of slip angle and a normalized camber stiffness of 0.0211 [1/o] and a twisting torque stiffness of 0.009 [m/o] at 10o of camber angle. That would be about 18 lb and 1.5 ft-lb, at 1o of slip angle 1.9 lb and 2.65 ft-lb at 10o of camber angle, respectively.

It is not as easy to extract as much detail for motorcycle tires from Cossalter's book, but for an "Sport 120/70/17" presumably under "normal conditions" on the same disk in Padua he shows a normalized cornering stiffness coefficient of 0.2 and a normalized aligning torque stiffness of 0.007 at 1o of slip and normalized camber stiffness coefficient of 0.2 and a normalized twisting torque stiffness 0.004 for the same tire at 10o of camber. That would be about 67 lb and 7.7 ft-lb, at 1o of slip angle 6.7 lb and 4.4 ft-lb at 10o of camber angle, respectively.

Oh. Sorry about ignoring the 30o and 45o part of your request. Doria only measured to 25o and shows camber force and twisting torque both increasing approximately linearly with camber angle. Cossalter shows up to 50o with approximately the same linearity.

For comparison purposes, consider the example Cossalter presents on pages 295-296 for "Gyroscopic effects generated by roll motion". The parameters he starts with are:
- roll rate = 0.5 radians/sec (28.65o/s)
- front wheel polar moment of inertia = 0.6 kgm2
- wheel spin rate = 100 radians/sec (71.6 mph with a 12.6 in. wheel)
- rake angle = 25o

Using steer torque = moment of inertia x spin rate x roll rate x cosine(rake angle), he calculates a torque about the steering axis of 27.2 N-m (20.0 ft-lb)

Meanwhile, the normalized twisting torque generated in the front tire contact patch of a 120-70-17 tire at 45o is 0.024 m. With a 289 lb vertical load on the front tire, that translates into a torque of 22.75 ft-lb in the same direction. About the 25o steering axis, that reduces to 20.6 ft-lb.

So the magnitudes are very similar. The difference, of course, is that the steering torque due to the gyroscopic effect varies with the roll rate and so can be near its maximum almost immediately, and the steering torque due to camber angle varies with the camber angle and so starts out at zero and builds nearly linearly to its maximum at the maximum lean angle.

[edited to expand answer to Scott's question about magnitudes in US units - AED]

Edited by - Andrew Dressel on 07/29/2013 12:36 PM
Go to Top of Page

Andrew Dressel
Male Standard Member
244 Posts


Milwaukee, WI
USA

Moto Guzzi

California Special

Posted - 07/28/2013 :  9:27 PM
quote:
Originally posted by James R. Davis
'Camber thrust' based on the consequences of the profile of the tire is what I have talked about here and elsewhere on the site. It does NOT depend on tire deformity. 'Camber thrust' as now being discussed not only depends on the profile of the tire, it depends on weight, lean angle, tire pressure, tire construction (biased vs radial), tire compounds, and road surface.



I don't follow you here. If there is no deformation of the tire, then the contact patch is a point, whether cambered or not. In that case, there are not two different distances from the inner and outer edges of that new contact patch to the wheel axle, so there can be no moment about a vertical axis generated in the contact patch.

Do you mean instead no deformation of the tire other than flattening in the contact patch?
Go to Top of Page

James R. Davis
Male Administrator
17286 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 07/28/2013 :  9:30 PM Follow poster on Twitter  Join poster on Facebook as Friend  
quote:
Do you mean instead no deformation of the tire other than flattening in the contact patch?


I do, indeed.
Go to Top of Page

Andrew Dressel
Male Standard Member
244 Posts


Milwaukee, WI
USA

Moto Guzzi

California Special

Posted - 07/29/2013 :  1:41 PM
Since some points have been spread out over many posts, let me attempt to summarize what has been asked and answered so far:

1. "There are TWO phenomena that are both referred to as 'camber thrust'."

One is a force, also called 'camber force' parallel to the plane of the contact patch, and perpendicular to the long axis of the contact patch. This is the phenomenon described in the Wikipedia article, Cossalter's Motorcycle Dynamics, and Pacejka's Tire and Vehicle Dynamics book. It requires deformation of the tire in the contact patch in addition to the flattening, and it may be large enough to produce all of the centripetal force required of the tire in a turn or more.

The other is a torque (or moment if you prefer), about an axis perpendicular to the plane of the contact patch. This is the phenomenon described in James' camber thrust glossary entry and James' tip 48. Cossalter calls it 'twisting torque'. It appears not to require more deformation of the tire in the contact patch than just the flattening, tends to turn the front wheel in the direction that it is leaned, and it may be as large as the torque about the steering axis due to the gyroscopic effect caused by cambering the spinning front wheel.

2. How does camber thrust play its roll in turning in these examples of thinner tyres with little width?

As measured in the laboratory, both the force and the torque due to rolling a cambered tire forward occur in even the narrowest bicycle tires. The magnitude of the force is also found to be approximately sufficient to provide all the centripetal force necessary. The magnitude of the torque, relative to other torques (caused by gyroscopic effects, trail, mass distribution, etc.) that contribute to turning the front wheel into a lean is not clear.

3. How far do you have to lean a bike over for camber thrust to start working?

In the testing by Cossalter and others, both the force and the torque appear to grow in magnitude nearly linearly with camber angle. Thus, the more you lean, the more you get.

4. How much force is involved?

The force can be estimated by simply multiplying the vertical load on the tire by the tangent of the camber angle.

The torque in one example provided by Cossalter was calculated for a 120-70-17 front tire at 45o of camber with a 289 lb vertical load about a 25o steering axis to be about 20 ft-lb.

5. [Other odds and ends

A "bike" can be self-stable with rigid wheels that have nearly a point contact patch and so do not generate force nor torque due to camber: https://www.youtube.com/watch?v=BK7QMJf0cv8

A "bike" can be self-stable with point contact patches, without gyroscopic effects, and without trail: https://www.youtube.com/watch?v=LbeENKn5kZU

[Edited immediately to spell the possessive form of James' name correctly, add some examples, fix formatting, and add synonyms. -AED]

Edited by - Andrew Dressel on 07/29/2013 3:07 PM
Go to Top of Page

James R. Davis
Male Administrator
17286 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 07/29/2013 :  2:07 PM Follow poster on Twitter  Join poster on Facebook as Friend  
Thank you - that was very clear.

I think readers here are still having a problem with the idea of camber thrust aiding the front tire in a turn. That is, providing turning torque.

What I described as camber thrust (the rolling coffee cup effect) does provide that assistance and I have described it as very weak - sufficient to make turning feel 'smoother', but not significant enough that it could be thought of as necessary to make a turn.

What you (and Cossalter, et al) describe as camber thrust does not provide a torque or turning moment. Instead, it is a source of centripetal force - that, along with gravity, determines lean angle and attempts to keep the motorcycle stable while in a turn.

Is that a fair summary at a grass roots level?
Go to Top of Page

Andrew Dressel
Male Standard Member
244 Posts


Milwaukee, WI
USA

Moto Guzzi

California Special

Posted - 07/29/2013 :  3:08 PM
quote:
Originally posted by James R. Davis

What I described as camber thrust (the rolling coffee cup effect) does provide that assistance and I have described it as very weak - sufficient to make turning feel 'smoother', but not significant enough that it could be thought of as necessary to make a turn.

What you (and Cossalter, et al) describe as camber thrust does not provide a torque or turning moment. Instead, it is a source of centripetal force - that, along with gravity, determines lean angle and attempts to keep the motorcycle stable while in a turn.

Is that a fair summary at a grass roots level?



I believe that is correct.
Go to Top of Page

scottrnelson
Advanced Member
6888 Posts
[Mentor]


Pleasanton, CA
USA

KTM

990 Adv, XR650L

Posted - 07/29/2013 :  4:11 PM
Thanks for clarifying this.

I now realize that the reason I was confused was because there were two different definitions. (I'm not saying that I totally understand it all yet, but it's much clearer than it was.)
Go to Top of Page

JanK
Male Junior Member
76 Posts


Ljubljana, Ljubljana
Slovenia

BMW

F650CS

Posted - 07/30/2013 :  2:55 AM
Not trying to confuse anyone, but after reading Cossalter, I'd have to say that the coffee cup analogy does not explain camber torque as described in the "camber thrust" glossary entry. In other words, the difference between the two radii is not in itself the reason for camber torque.

The thing is that a rolling cup, once it's in a steady state circular motion, experiences no camber torque at all. If it did, it would either have to accelerate or the torque would have to be balanced by another torque.

Rather than try to fumble my way through an explanation, I've grabbed two pages from Amazon's preview of Cossalter's book:




In summary:
- Tire's turn centre is always displaced relative to the intersection of tire's axis with the road plane. Thus tire always generates some small amount of camber torque. In a steady state turn this torque is compensated by other torques around the steering head axis to produce a zero net torque.
- Coffee cup's turn centre always coincides with the intersection of cup's axis with the road plane. Thus coffee cup never generates camber torque.
Go to Top of Page
Page: of 3 Previous Topic Discussion Topic Next Topic  
Previous Page | Next Page
Jump To:
All Things (Safety Oriented) Motorcycle © Master Strategy Group Go To Top Of Page
  This page was generated in 0.47 seconds. Powered By: Snitz Forums 2000 Version 3.4.05