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Automobile rear wheel travel 3
- Thread starter CoryPad
- Start date
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1
- #2
GregLocock
Automotive
Don't know, but for a large sedan with a good ride and very good handling it needs 100 mm into jounce and 120-135 into rebound. For a road going car on smooth roads you can get away with 50 mm up and 30 mm down, with some (severe) compromises.
Cheers
Greg Locock
Cheers
Greg Locock
hi. i think that u can find ur answer inthis sit: M.NOROOZI
ENG.DIVISION
ENG.DIVISION
- Thread starter
- #4
noroozi,
Your link does not work. I also tried but that requires a login which I don't have. Thanks anyway.
Your link does not work. I also tried but that requires a login which I don't have. Thanks anyway.
235mm bottom to top? That sounds like an awefull lot. Even 135 sounds like alot for most cars. Heck, I don't even think a stock ranger has over 200. For one of those tiny cars i'd only think 120 or 130 total, but i'm just guessing on all of it.
-=Whittey=-
-=Whittey=-
GregLocock
Automotive
The reason you might need more rebound than jounce is that it is very easy for the inside rear wheel to lift on a front wheel drive car. This imposes a significant non-linear effect, just when it is least helpful, ie when you are tearing around a corner. The deflections I quoted are metal to metal, you can figure on never reaching that jounce limit as the snubbers cut in, even in durability type events, where we usually work to compressing conventional snubbers to 1/3 of their original length. This typically knocks 20 mm or so off the jounce articulation.
Cheers
Greg Locock
Cheers
Greg Locock
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2
- #10
CoryPad,
Fundamentally, a suspension systems static deflection (dst) is equal to the ratio of the corner weight (Mcg) to the suspension spring rate (ks) measured at the wheel. Back solve from the formula fn = (ks/ Mc)-2/(2pi) to get the static deflection.
For a 1 Hz ride frequency (typical) the static deflection would be about 250mm. Notice you don’t need to know the mass or suspension spring rate to get to this point. Static deflection is only a factor of desired ride frequency.
Now, the deflection that a suspension must be able to allow is determined the same way. It depends on how large an acceleration event you wish to absorb. To contain a “bump” of 0.5 g, about 125mm of travel is needed at a 1Hz frequency. This is all without damping and ignoring all the non-linearities and the components that are needed to slow the unsprung mass down at the ends of the available travel. These components can bump-stops, extension stops (springs or hydraulic), non-linear rate springs, etc. You can get by with less travel, you just need to attenuate the motion before you hit a hard stop.
I hope this helps.
Best regards,
Matthew Ian Loew
Sr. Engineer
Fundamentally, a suspension systems static deflection (dst) is equal to the ratio of the corner weight (Mcg) to the suspension spring rate (ks) measured at the wheel. Back solve from the formula fn = (ks/ Mc)-2/(2pi) to get the static deflection.
For a 1 Hz ride frequency (typical) the static deflection would be about 250mm. Notice you don’t need to know the mass or suspension spring rate to get to this point. Static deflection is only a factor of desired ride frequency.
Now, the deflection that a suspension must be able to allow is determined the same way. It depends on how large an acceleration event you wish to absorb. To contain a “bump” of 0.5 g, about 125mm of travel is needed at a 1Hz frequency. This is all without damping and ignoring all the non-linearities and the components that are needed to slow the unsprung mass down at the ends of the available travel. These components can bump-stops, extension stops (springs or hydraulic), non-linear rate springs, etc. You can get by with less travel, you just need to attenuate the motion before you hit a hard stop.
I hope this helps.
Best regards,
Matthew Ian Loew
Sr. Engineer
GregLocock
Automotive
vanb
Your equation is probably correct, but all it does, as Matthew the senior engineer shows, is look at the spring length change between zero load and laden, for a given ride frequency. This is not a very helpful figure in practice - suspensions are usually designed so that at full droop the spring is still compressed (think about why we do that).
Matthew wrote "Now, the deflection that a suspension must be able to allow is determined the same way. " Um, no, unfortunately large real world inputs are more likely to be square edges than sine waves, so a simple harmonic analysis will fail. You don't see sinewaves like you are proposing on the road, except at a test track. Damping would not affect that solution.
Cheers
Greg Locock
Your equation is probably correct, but all it does, as Matthew the senior engineer shows, is look at the spring length change between zero load and laden, for a given ride frequency. This is not a very helpful figure in practice - suspensions are usually designed so that at full droop the spring is still compressed (think about why we do that).
Matthew wrote "Now, the deflection that a suspension must be able to allow is determined the same way. " Um, no, unfortunately large real world inputs are more likely to be square edges than sine waves, so a simple harmonic analysis will fail. You don't see sinewaves like you are proposing on the road, except at a test track. Damping would not affect that solution.
Cheers
Greg Locock
Greg,
I agree with you. The calculations I presented represent a first pass analysis of what may be required to determine the upper bounds of travel. If you know of a better way to calculate the amount of travel that would be required under certain design considerations it would benefit this thread. More travel is generally better.
It indeed does break down pretty rapidly under the conditions that I mentioned (all the additional damping and spring mechanisms in the suspension system) and you followed up on make the equations presented invalid at the travel extremes. The necessity to limit extension travel to prevent the spring from going beyond the free length is especially valid. Leaf springs and air springs can result in greater effective free lengths, however. But these springs are non-linear as well, so do not help us in identifying a better fundamental equation.
Best regards,
Matthew Ian Loew
Sr. Engineer
I agree with you. The calculations I presented represent a first pass analysis of what may be required to determine the upper bounds of travel. If you know of a better way to calculate the amount of travel that would be required under certain design considerations it would benefit this thread. More travel is generally better.
It indeed does break down pretty rapidly under the conditions that I mentioned (all the additional damping and spring mechanisms in the suspension system) and you followed up on make the equations presented invalid at the travel extremes. The necessity to limit extension travel to prevent the spring from going beyond the free length is especially valid. Leaf springs and air springs can result in greater effective free lengths, however. But these springs are non-linear as well, so do not help us in identifying a better fundamental equation.
Best regards,
Matthew Ian Loew
Sr. Engineer
GregLocock
Automotive
Well I suppose the easy answer is that you define a road profile and a speed at which you wish to avoid bottoming out, then build a model to select appropriate spring and damper rates.
That's the easy answer, and it works quite well for race vehicles and other cases where 10-20 year life is not required.
For real cars in practice you (a) come up with a set of load cases, and then (b) hack together a suspension geometry (including bump stops) and then (c) check that the body structure doesn't fall apart.
Then you stuff around with everything until it works.
So here's a rule of thumb for a road going car:
Jounce travel (metal to metal) : at least 50 and preferably 100 mm
Rebound travel: Ideally sufficient that in a 1g corner all four wheels are still on the road.
Of course the problem is that your worst input, kerb strike or chuck hole, is reacted by lifting the car body as well as deflecting the suspension, so the analysis gets multi dimensional very quickly.
This also ignores the way that you set the road spring rates. So far as I am aware the best way to do this as a starting point is to say that the rear axle frequency should be a certain multiple of the front axle frequency, and that the front axle frequency should be between 1 and 2 Hz, for a normal car. 1 for luxury 2 for sports. f=1/(2*pi)*sqrt(spring stiffness referred to wheel centre/sprung mass at that wheel). Note that very few geometries give you the same spring rate at the wheel centre as the actual spring rate, typical values are .6 (appalling) to 0.9. A kinematic diagram of the suspension can be used to determine this.
The front to rear ratio is set by the 'flat ride' criterion, whereby the pitch response to a step input is minimised at a selected speed by setting the rear frequency slightly higher than the front's. A 2 DoF model is necessary to get this right. Typically Ffront=1.2 Hz, Frear =1.5 Hz.
Having said that this criterion is pretty much irrelevant for sports versions of road going cars. Cheers
Greg Locock
That's the easy answer, and it works quite well for race vehicles and other cases where 10-20 year life is not required.
For real cars in practice you (a) come up with a set of load cases, and then (b) hack together a suspension geometry (including bump stops) and then (c) check that the body structure doesn't fall apart.
Then you stuff around with everything until it works.
So here's a rule of thumb for a road going car:
Jounce travel (metal to metal) : at least 50 and preferably 100 mm
Rebound travel: Ideally sufficient that in a 1g corner all four wheels are still on the road.
Of course the problem is that your worst input, kerb strike or chuck hole, is reacted by lifting the car body as well as deflecting the suspension, so the analysis gets multi dimensional very quickly.
This also ignores the way that you set the road spring rates. So far as I am aware the best way to do this as a starting point is to say that the rear axle frequency should be a certain multiple of the front axle frequency, and that the front axle frequency should be between 1 and 2 Hz, for a normal car. 1 for luxury 2 for sports. f=1/(2*pi)*sqrt(spring stiffness referred to wheel centre/sprung mass at that wheel). Note that very few geometries give you the same spring rate at the wheel centre as the actual spring rate, typical values are .6 (appalling) to 0.9. A kinematic diagram of the suspension can be used to determine this.
The front to rear ratio is set by the 'flat ride' criterion, whereby the pitch response to a step input is minimised at a selected speed by setting the rear frequency slightly higher than the front's. A 2 DoF model is necessary to get this right. Typically Ffront=1.2 Hz, Frear =1.5 Hz.
Having said that this criterion is pretty much irrelevant for sports versions of road going cars. Cheers
Greg Locock
Greg,
Nice post. Just a couple remarks:
Strut type suspensions allows a 1:1 motion ratio between the wheel and the spring/damper. One of their few good points.
quote: Rebound travel: Ideally sufficient that in a 1g corner all four wheels are still on the road.
This would rule out most VWs from the past few decades!![[smile]](/data/assets/smilies/smile.gif "[smile] [smile]")
Best regards,
Matthew Ian Loew
m.loew@ssss.com
Nice post. Just a couple remarks:
Strut type suspensions allows a 1:1 motion ratio between the wheel and the spring/damper. One of their few good points.
quote: Rebound travel: Ideally sufficient that in a 1g corner all four wheels are still on the road.
This would rule out most VWs from the past few decades!
Best regards,
Matthew Ian Loew
m.loew@ssss.com
GregLocock
Automotive
You are right, I'd been thinking too much about velocity diagrams and not enough about cheap simple suspensions - for some reason I never get to put a McPherson or Chapman strut model together! For my sins I've just been advised that that's the next project. Actually very few struts achieve 1:1 because the packaging to get the spring moving as much as the wheel does vertically is so bad, tho I think the Honda CRV rear suspension may do it. Our base car still has a beam axle, which of course has a 1:1 motion ratio in two wheel bounce, rather less in roll.
I've just thumbed through Milliken squared and he has done the math:
Kwheel=Kspring*IR^2+Fspring*(dIR/dz)
where IR is the motion ratio, and you can ignore the second term for a simple analysis. Cheers
Greg Locock
I've just thumbed through Milliken squared and he has done the math:
Kwheel=Kspring*IR^2+Fspring*(dIR/dz)
where IR is the motion ratio, and you can ignore the second term for a simple analysis. Cheers
Greg Locock
All,
I know we are getting off of Cory’s request for specific information on small car rear suspension travel, but I went back through my files and found a good reference for calculation of required travels. The SAE Technical Paper 983065 Development of “Camion” Truck Winner at ‘97 Dakar Rally chronicles Hino’s efforts to produce a winning truck for the Dakar Rally-Raid. Of specific interest is the parametric study used to determine the appropriate stroke, spring rate, and damping. Pretty much what Greg indicated in his first sentence above.
Hope this helps! Best regards,
Matthew Ian Loew
m.loew@ssss.com
I know we are getting off of Cory’s request for specific information on small car rear suspension travel, but I went back through my files and found a good reference for calculation of required travels. The SAE Technical Paper 983065 Development of “Camion” Truck Winner at ‘97 Dakar Rally chronicles Hino’s efforts to produce a winning truck for the Dakar Rally-Raid. Of specific interest is the parametric study used to determine the appropriate stroke, spring rate, and damping. Pretty much what Greg indicated in his first sentence above.
Hope this helps! Best regards,
Matthew Ian Loew
m.loew@ssss.com
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