bump acceleration question

[CU Matt]

I have read in How to Make Your Car Handle that it is common to apply a 5g bump acceleration when designing suspension linkages. In previous forum posts, I have seen people apply this by stating that the normal force on a tire is 5 times as much as its static normal force. I have also read that the 5g acceleration is commonly interpreted as if an accelerometer on the wheel registered 5 g's. These two statements seem contradictory to me, so I was looking for some clarification in my understanding.

If I assume a 5g acceleration and take the sum of the forces in the vertical direction, I find the following:

Sum of Forces=ma
0g bump static corner weight(W) - 0g bump static normal force(N) +force due to bump(Fb)=ma

W-N+Fb=ma

Since W-N=0, then
Fb=ma=5mg

and the normal force on the tire with the 5g bump (Nb) would then be:

Nb=N+Fb=6mg

Intuitively this also makes sense because if I am standing on the ground, with no acceleration, then the normal force on me is equal to my weight. If I experience a bump acceleration of 1g then the normal force at my feet would be twice my weight.

Consequently, it seems to me that when people say that they have applied a bump acceleration of xg's, they are really applying a bump acceleration that yields a normal force of (x+1)mg.

I understand that it is questionable whether or not the dynamics of a bump will ever actually allow the normal force to reach a normal force found by the analysis above, but such is not my question. I am wondering if applying a 5g acceleration actually means a 5g acceleration on the tire, or means that the normal force under a bump acceleration will be equal to 5mg, but a accelerometer on the tire would register 4g's. Thanks.
-Matt

Puhn, F., How to Make Your Car Handle, The Berkley Publishing Group, 1981

[CU Matt]

I have read in How to Make Your Car Handle that it is common to apply a 5g bump acceleration when designing suspension linkages. In previous forum posts, I have seen people apply this by stating that the normal force on a tire is 5 times as much as its static normal force. I have also read that the 5g acceleration is commonly interpreted as if an accelerometer on the wheel registered 5 g's. These two statements seem contradictory to me, so I was looking for some clarification in my understanding.

If I assume a 5g acceleration and take the sum of the forces in the vertical direction, I find the following:

Sum of Forces=ma
0g bump static corner weight(W) - 0g bump static normal force(N) +force due to bump(Fb)=ma

W-N+Fb=ma

Since W-N=0, then
Fb=ma=5mg

and the normal force on the tire with the 5g bump (Nb) would then be:

Nb=N+Fb=6mg

Intuitively this also makes sense because if I am standing on the ground, with no acceleration, then the normal force on me is equal to my weight. If I experience a bump acceleration of 1g then the normal force at my feet would be twice my weight.

Consequently, it seems to me that when people say that they have applied a bump acceleration of xg's, they are really applying a bump acceleration that yields a normal force of (x+1)mg.

I understand that it is questionable whether or not the dynamics of a bump will ever actually allow the normal force to reach a normal force found by the analysis above, but such is not my question. I am wondering if applying a 5g acceleration actually means a 5g acceleration on the tire, or means that the normal force under a bump acceleration will be equal to 5mg, but a accelerometer on the tire would register 4g's. Thanks.
-Matt

Puhn, F., How to Make Your Car Handle, The Berkley Publishing Group, 1981

[Jersey Tom]

Depends on how you zero your sensors. Some people like to say Az = 0 G at stand still.. some like to say it's 1 or -1 G. Either way works. Personally it makes sense to zero the sensors at 0.0G.

Either way, doesn't matter in this case since using 5G bump acceleration is an arbitrary value.

[CU Matt]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Jersey Tom:
Depends on how you zero your sensors. Some people like to say Az = 0 G at stand still.. some like to say it's 1 or -1 G. Either way works. Personally it makes sense to zero the sensors at 0.0G.

Either way, doesn't matter in this case since using 5G bump acceleration is an arbitrary value. </div></BLOCKQUOTE>

True, the 5g bump acceleration is arbitrary to a degree, but it is not completely arbitrary. Accounting for a 3 to 5g bump acceleration is fairly standard, so it is probably in the range that a bump acceleration would be.

Even if it was completely arbitrary, I would imagine that there is a standard to properly communicate the acceleration you tested for. I have read papers where the loading conditions of the tire are given, such as 1.4 g lateral acceleration and 5g bump and 1g acceleration, ect. These values are meaningless if there is not a standard by which they are communicated. If they are meaningless, then so be it; but that does not seem to be the case.

Now perhaps in the case of bump acceleration, the standard is to say Az=0 at stand still or 1, or -1G. If this is the case then it would be good for people to state the zero that they use, but I haven't seen people state the zero before.

[absolutepressure]

I don't think applying a 5g bump would be the best way to go about it. In my Dynamical Systems class, we modeled a quarter car, derived the system governing equations, and introduced a displacement step input (like going over a 90* bump like a curb, but not so high). From the response of the system, (which in this case will be displacement), you can get forces knowing your spring rate, and from that, acceleration. We also introduced a sign wave to simulate a wash board road, but I don't think that's necessary for fsae.

What I would recommend is do the step input with a bump you anticipate hitting (hopefully not more than 1.5in high or you'll have clearance issues, I bet!), then from that, your output will be the the response of your suspension. With the way you propose, you're setting an acceleration, that for all you know can only be generated with a 2in bump.

Don't constrain the outputs, constrain the inputs and find your outputs.

[CU Matt]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by absolutepressure:
I don't think applying a 5g bump would be the best way to go about it. In my Dynamical Systems class, we modeled a quarter car, derived the system governing equations, and introduced a displacement step input (like going over a 90* bump like a curb, but not so high). From the response of the system, (which in this case will be displacement), you can get forces knowing your spring rate, and from that, acceleration. We also introduced a sign wave to simulate a wash board road, but I don't think that's necessary for fsae.

What I would recommend is do the step input with a bump you anticipate hitting (hopefully not more than 1.5in high or you'll have clearance issues, I bet!), then from that, your output will be the the response of your suspension. With the way you propose, you're setting an acceleration, that for all you know can only be generated with a 2in bump.

Don't constrain the outputs, constrain the inputs and find your outputs. </div></BLOCKQUOTE>

I think your approach would be beneficial if the inputs could be known; however, I think knowing what the inputs are could be challenging. When a car is cornering, much of its damper/spring travel can get used up. Say you have 2 inches of spring/damper travel, a 1 to 1 motion ratio and your spring/damper unit compresses 3/4 of an inch during cornering and then you hit a 0.5 inch bump. Under this situation it is possible that the spring would bottom out and there would be increased forces that directly relate to the bump acceleration. It is also possible that the vehicle would roll such that the spring wouldn't bottom out. As a result it would be unclear what the inputs could be constrained as.

Really, to get a clearer picture of what bump accelerations are valid, I think I would need Data Aquisition results, but I presently don't have that.

While I am not sure applying a 5g bump would be the best way either, i'll probably end up going off of results from a 3g to 5g bump acceleration with 0 g's assumed at rest. Although it would be nice to know if other teams have any data aq results that have indicated what types of bump accelerations should be assumed.