cygnusx1 Posted December 7, 2006 Share Posted December 7, 2006 Dave, Theoretically, no. Practically, it could flex/bow the strut piston shaft. Quote Link to comment Share on other sites More sharing options...
JMortensen Posted December 7, 2006 Share Posted December 7, 2006 I think in practical application there is quite a bit of bushing deflection. Quote Link to comment Share on other sites More sharing options...
thehelix112 Posted December 7, 2006 Author Share Posted December 7, 2006 Ok I am still obviously not completely understanding this. In that case that the strut has positive castor, won't the twisting moment cause an effective change in length due to the vertical component of the end of the struts arc of movement? Damn I need a diagram. If that makes no sense I'll draw one. Dave Quote Link to comment Share on other sites More sharing options...
cygnusx1 Posted December 7, 2006 Share Posted December 7, 2006 You are really close to understanding this. You are thinking that the strut is angled forward so that a horizontal force at the top will make it compress or extend because of the angle. That is correct. What is incorrect is the assumption that the force is horizontal. It's not, it's always perpendicular to the strut axis so the angle between the force and the strut is 90, cancelling any lift/droop. Quote Link to comment Share on other sites More sharing options...
thehelix112 Posted December 7, 2006 Author Share Posted December 7, 2006 Not quite. I am thinking the strut is angled forwards, thus any force perpendicular to the strut axis will have a vertical component. Unless the car is going to move, this vertical component will be reacted by the strut tower and the suspension will compress. No? Dave PS. Thanks for bearing with me. Quote Link to comment Share on other sites More sharing options...
jt1 Posted December 7, 2006 Share Posted December 7, 2006 I think this is the Mark Ortiz article referred to by Cary. It's talking about wheel loading, but there are a lot of revelant points there. EFFECT OF CALIPER MOUNTING POSITION Mark Ortiz April 2004 Chassis Newsletter What effect on wheel loading does the positioning of the calipers in a leading or trailing location have – i.e. mounted at 3 and 9 o’clock positions? Does a trailing caliper add or subtract load on the front tires? In a rear independent suspension, does a leading caliper add or subtract wheel loading, and is it the same in a live axle situation? The short answer is no. Caliper location has no effect whatsoever on wheel loading. Having the caliper’s mass lower or higher does have a very minute effect, because it affects the CG location a tiny bit, but there is no difference between a 3 o’clock mounting position and a 9 o’clock position. However, there is an effect on bearing loads. It might seem counterintuitive that we can change the bearing loads and not change the tire loads, but that is in fact the case. As the questioner appears to have considered, the disc tries to carry the caliper upward if the caliper is trailing, and downward if the caliper is leading. That reduces bearing loads if the caliper is trailing, and increases bearing loads if the caliper is leading. However, these forces are reacted entirely within the hub/bearing/spindle/upright/caliper/disc/hat assembly, and do not change the loads on other parts of the car. We can think of it like this: Gravity acts downward on the car, with additions and subtractions due to inertia effects and aerodynamic effects. The road surface holds the car up. Or, we may say the road holds the tire up; the tire holds the wheel up; the wheel holds the hub up; the hub holds the bearings up; the bearings hold the spindle up; the spindle holds the upright up; the upright holds the suspension up; the suspension holds the sprung mass up. If the caliper exerts an upward force on the upright and a downward force on the disc, that just means the brake is helping the bearings and spindle hold the upright up. It doesn’t change the total support force, only the load path within some of the unsprung components. It is worth noting that in braking there are also horizontal forces acting through the wheel bearings. The car is trying to keep going forward at a constant speed. The road surface is exerting a rearward force on the car, through the tires, wheels, hubs, bearings, spindles, uprights, and suspension. We can reduce the bearing loads due to this component if we mount the caliper above center, or increase the bearing loads if we mount the caliper below center. In fact, the horizontal force may be greater than the vertical force on the tire. With racing slicks on dry pavement, the horizontal force may be 1.3 or more times as great as the vertical load on the tire. So for least bearing loads during braking, the caliper should be somewhere in the upper rear quadrant – around 1 o’clock or 11 o’clock, depending on which wheel we’re looking at, and from what direction. Now, do we actually want maximum cancellation of the bearing loads by the brakes? We might suppose so, but actually there is an argument for not having maximum cancellation. The effective radius of the brake (roughly the radius to the middle of the pad) is often less than half of the tire effective radius. This means that the force at the caliper is more than twice the rearward force at the tire contact patch, and it may also exceed the vector sum of the vertical and horizontal forces at the contact patch. Consequently, the caliper force may not only reduce the bearing loads, but reverse them. If there is any free play in the bearings, or deflection in the components, this load reversal may result in a vibration or a small variation in the steer angle of the wheel. So there is a case for building the components nice and strong, and positioning the calipers so the bearing loads will not reverse. Of course, as a practical matter, if we are using purchased calipers we need to mount them with the bleed screws at the top, or very nearly so, just to facilitate good brake bleeding without requiring the calipers to be dismounted. This may well outweigh any theoretical considerations. If we are designing from a blank sheet of paper, we don’t face this constraint, but most of us, most of the time, are designing around purchased calipers. Another practical constraint is packaging, particularly of the steering arms and cooling ducts. There are some ways in which we can affect wheel loads by the design of the brake system and the suspension. I am referring here to the longitudinal “anti†or “pro†effects: anti-dive or pro-dive in the front suspension, anti-lift or pro-lift at the rear. With independent suspension, it makes a difference to these effects whether the brakes are inboard or outboard. With a beam axle, it makes a difference if the calipers are mounted directly to the axle, or on birdcages or floaters that rotate on the axle and have their own linkages. However, with all of these, we cannot significantly alter the loading on the front or rear wheel pair, nor on all four wheels. We can change the way the sprung mass moves in response to braking, and this may have small effects on CG height, with corresponding small effects on overall load transfer. But the big effects come from having geometry differences on the right and left sides of the car. These may be present even in supposedly symmetrical road racing cars, because no car stays symmetrical when it rolls. In oval track cars, we often design in, or adjust in, asymmetry even in the static condition. Such asymmetry can produce significant changes in diagonal percentage when braking, and we can use these to tune corner entry behavior. All such effects are independent of the “clock†position of the caliper mount. I'm still thinking about it. I fully agree caliper position doesn't affect the chassis, but I'm not convinced it can't compress the strut. John Quote Link to comment Share on other sites More sharing options...
jt1 Posted December 7, 2006 Share Posted December 7, 2006 posted by cygnusx1 The wheel wants to pivot around the brake pads when squeezed, but it can't because the spindle holds it in place. The force of the wheel against the spindle exactly counteracts the force of the caliper. This cancellation happens all inside the wheel-struthousing-caliper relationship. The strut itself (the piston) sees none of what is going on. OK, so my vertical force exerted by the 3:00 caliper is not resisted by the unibody on top of the strut (which I thought might compress the strut), but rather by a downward force on the spindle (not compressing the strut, only deflecting the spindle/caliper mount an insignificant amount). Good explanation, Thanks. I've enjoyed this discussion!!!! John Quote Link to comment Share on other sites More sharing options...
cygnusx1 Posted December 8, 2006 Share Posted December 8, 2006 Not quite. I am thinking the strut is angled forwards, thus any force perpendicular to the strut axis will have a vertical component. Unless the car is going to move, this vertical component will be reacted by the strut tower and the suspension will compress. No? Dave PS. Thanks for bearing with me. I have been thinking about this all night after I understood what you are seeing. What you are describing is the strut assembly trying to rotate forward in the direction of the wheels rotation from braking induced moments. With positive caster, this force does have a vertical component and will want to lift the front of the car under braking. However, this is a moment about the ball joint, produced by a force on the contact patch opposite the direction of travel of the car. This is not at all related to the clock position of the caliper. Whew. My brain hurts. Quote Link to comment Share on other sites More sharing options...
johnc Posted December 8, 2006 Share Posted December 8, 2006 Sorry, late to the part, cygnusx1 is right. Also, where the caliper is clocked has little to no affect on the sprung mass and little if any affect on the unsprung mass. I tested the position of the rear calipers on the ROD and could not discern any difference in handling or corner weights. Quote Link to comment Share on other sites More sharing options...
thehelix112 Posted December 8, 2006 Author Share Posted December 8, 2006 I am glad we all agree that the position of the caliper is irrelevant to its longitudinal effect on suspension. Now back to my original post on the lateral effect; a caliper that has been clocked lower will lower the centre of mass of the strut assembly. I believe this will affect how the lateral force from the contact patch gets translated to a moment on the strut assembly. Ie, if the centre of mass is exactly at the stub axle (spindle) level then no moment will occur and the force will be transferred directly to the ball-joint/LCA and strut tower. If the centre of mass is above the spindle, there will be a moment attempting to rotate the strut top in the opposite direction of the turn. For example, the car is turning left, therefore the contact patch/wheel/spindle are exerting a force left. This force is below the spindle, so the strut will see a moment that is attempting to move the strut top right, and the ball-joint left. Once again I fear I need to either re-write that, or draw a diagram, neither of which I have time for. I hope you get the drift of what I am asking (for clarification on). Dave PS. I realise there are many more significant things to consider when tuning a car, and that considering this is perhaps less than supremely practical. I still want to understand whats going on more fully. Quote Link to comment Share on other sites More sharing options...
cygnusx1 Posted December 8, 2006 Share Posted December 8, 2006 Started this thread to stop more of a hi-jack from this thread: http://forums.hybridz.org/showpost.php?p=712084&postcount=28 In response to I am not sure what you mean about mounting the calipers `low on the spindles'. Are you just talking about the 5/7 o'clock mounting position? Some random thoughts: while you are correct in that this lowers the centre of gravity of the strut assembly, as this is sprung, it does not affect the centre of gravity of the car at all; and thus its only affect is in determining the magnitude of the moment placed on the chassis by the strut assembly when in a turn. This moment on the inside wheel will attempt to pitch the car undesirably (out of the corner), but on the outside wheel it will attempt to pitch the car into the corner. I would hazard a guess that these two moments would all-but cancel each other out, and you're left with no impact at all. Unless I have missed something incredibly obvious. This is something I have never completely understood, and am keen to do so. Though it might be better in another thread? Dave Back to the original point....completely negligable forces. You would have more positive effect on overall CG by removing the cigarette lighter. Quote Link to comment Share on other sites More sharing options...
bjhines Posted December 8, 2006 Share Posted December 8, 2006 except for the wheel bearing loading point... hehehe... Quote Link to comment Share on other sites More sharing options...
thehelix112 Posted December 8, 2006 Author Share Posted December 8, 2006 cygnusx, Acknowledged, but is that the way the forces would be applied? Dave Quote Link to comment Share on other sites More sharing options...
tube80z Posted December 8, 2006 Share Posted December 8, 2006 Now back to my original post on the lateral effect; a caliper that has been clocked lower will lower the centre of mass of the strut assembly. I believe this will affect how the lateral force from the contact patch gets translated to a moment on the strut assembly. I'm not quite following you here. I get that the suspended mass and non-suspended masses can have seperate CGs and this has an effect on suspension kinematics. But, when we measure a car for CG we lock the suspension in place and tilt it to find the angle it will balance at, which is more accurate than the method of raising one end and measuring the change in scaled weight. In this scenario anything lowered will reduce the CG of the system as a whole. The lateral weight transfer is additive of the suspended mass and non-suspended masses. Both are in the same direction. And at the end of the day we're looking for less of this to maximize grip. Cary Quote Link to comment Share on other sites More sharing options...
thehelix112 Posted December 9, 2006 Author Share Posted December 9, 2006 Cary, What I am talking about is how the force applied by the tyre affects the strut assembly, and then distinctly, how this affected strut assembly affects the chassis. Dave Quote Link to comment Share on other sites More sharing options...
tube80z Posted December 9, 2006 Share Posted December 9, 2006 What I am talking about is how the force applied by the tyre affects the strut assembly, and then distinctly, how this affected strut assembly affects the chassis Okay, but isn't that the same as talking about suspension kinematics, or in this case the side view geometry? I think there's some subtlety that you see that I'm not picking up on. You can test compliance by pulling on wheel pairs and you need to lock down the suspension or the susension will move. In the case of lateral testing the car tries to raise (I should point out I've only done one car so beware of the notorious data point of 1). Is that more like what your seeing? cary Quote Link to comment Share on other sites More sharing options...
thehelix112 Posted December 11, 2006 Author Share Posted December 11, 2006 Cary, The topic does clarify what I am interested in, and yes, I am sure it will have an effect on suspension, but thats not what I am primarily concerned with. I am not talking about compliance. I am talking about how (note: not how much, I know it will be insignificant) the forces are applied from the tyre to the chassis, and how they change with caliper mounting position. Does that help? Dave Quote Link to comment Share on other sites More sharing options...
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