Assumed Lateral and Longitudinal Acceleration for 1st year team.

[OspreysGoSWOOP]

Hello all,

1st Year team here.

I'd like to get some input from experienced teams on what types of lateral and longitudinal acceleration your car experiences.

I've read a ton of posts and I've seen assumptions anywhere from 1g to 3g's (for both lat and long). Using numbers from some team's acceleration event times I came up with average .75g longitudinal acceleration.

What did you assume or find via testing? Thanks.

[OspreysGoSWOOP]

Hello all,

1st Year team here.

I'd like to get some input from experienced teams on what types of lateral and longitudinal acceleration your car experiences.

I've read a ton of posts and I've seen assumptions anywhere from 1g to 3g's (for both lat and long). Using numbers from some team's acceleration event times I came up with average .75g longitudinal acceleration.

What did you assume or find via testing? Thanks.

[BMH]

Chapter 3 of Carrol Smith's Tune To Win should help give you some numbers for this.

[exFSAE]

What can you figure out from skid pad times?

[OspreysGoSWOOP]

Should I be using skid pad times/radius for lateral load? I used 45mph through a 7.25m turn.

I calculated 7.25m from autocross minimum hairpin turn radius of 9m (Rule D7.2.1) minus center of minimum track width
r=9-1.75=7.25m

Maybe I should be using r=15.25/2 + 1.5 = 9.125m

[Mikey Antonakakis]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by OspreysGoSWOOP:
Should I be using skid pad times/radius for lateral load? I used 45mph through a 7.25m turn.

I calculated 7.25m from autocross minimum hairpin turn radius of 9m (Rule D7.2.1) minus center of minimum track width
r=9-1.75=7.25m

Maybe I should be using r=15.25/2 + 1.5 = 9.125m </div></BLOCKQUOTE>
Skidpad times would be a good place to start. Looks like you've already found the skidpad radius too, so go for it. Check out Rule D6.8.1.

[Mike Cook]

1.5 g lateral
1 g long.

[Wagz]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by OspreysGoSWOOP:
Should I be using skid pad times/radius for lateral load? I used 45mph through a 7.25m turn.

I calculated 7.25m from autocross minimum hairpin turn radius of 9m (Rule D7.2.1) minus center of minimum track width
r=9-1.75=7.25m

Maybe I should be using r=15.25/2 + 1.5 = 9.125m </div></BLOCKQUOTE>

It won't matter what radius you use if you don't have the correct speed. What kind of lateral accelerations did 45 mph through a 7.25 m turn give you? I would kill for our car to be able to do something like that. Since you don't seem to know about the vehicle speeds through the corners, you were pointed to skidpad times. The acceleration can be determined for the fixed skidpad layout if you can somehow find the times of the cars that go around it, hmmm....

[thewoundedsoldier]

For some reason max braking grip always seems higher than max acceleration grip. And I don't buy the whole aero thing, myself.

@ Mike - I love it! everytime I see a bunch of people yacking out longer answers i know you have a two- or three- word reply coming.

[The AFX Master]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by thewoundedsoldier:
For some reason max braking grip always seems higher than max acceleration grip.
</div></BLOCKQUOTE>

Easy.. 4 wheels stop the car, 2 wheels give thrust to it.

[OspreysGoSWOOP]

Thank you everyone for your help. I now have reasonable assumptions. In order to make the forums more useful, I'll post my work below.


Longitudinal
d=75m
t=4.5s (min 3.97, max 5.2)
v=d/t = 16.67m/s or 37mph
a=v/t = 16.67 / 4.5 = .38g
FOS 2.5: a = .95 or 1g

Lateral
r=(15.25+3)/2 = 9.125m (assumes perfect driving in center of skid pad circle)
OR
r=17.1/2 = 8.55m (Rule D6.8.1)
I used r = 8.55m
d=2*pi*r = 57m
t=5.2s
v=d/t= 11m/s or 25mph
a = v^2/r = 16.67 / 4.5 = 1.36g
FOS 2.5: a = 3.4g

Cheers and thank you all.

[MRT_Xii]

Steady state assumption will work for lateral values as you are using skid pad times, which is essentially steady state. Do you think this holds for acceleration time? I would be more concerned with breaking than acceleration anyways, and there is a fairly good hint why on this very page...

2.5 FOS is a little excessive... Hell, your car will flip much before 3.4 lateral gsss.

[exFSAE]

2.5 FOS is pretty insane, dude. Even 1.2 is a really damn high FOS

[OspreysGoSWOOP]

Excellent points. I just picked 2.5 out of thin air.

Thinking about an FSAE car flipping reminded me of this toy.
http://www.youtube.com/watch?v=ypD7k_UK4CY#t=0m10s

Someone should make an FSAE car that can do that. http://fsae.com/groupee_common/emoticons/icon_biggrin.gif

[Mike Cook]

If you're a first year team its better to error on the side of too strong. Your numbers look correct for skid pad, but in a transient maneuver, you might see a good deal more lateral force. For longitudinal, during a launch, we generally see around 1g or higher, so averaging acceleration over 75m is a really bad way of finding peak accel because it isn't a steady state event.

Braking can be quite high as well. 1.5g or more is reasonable for braking.

When you design your suspension, most failures in linkages are due to buckling. Compression loads will buckle your tubes. Repeat it. Repeat it. If you initialize buckling by putting your pushrod load into your lower a-arm, you will buckle your tube much earlier than Euler's critical buckling load. For links in bending and compression, it is almost always better to use the biggest thinnest tube possible. For example, instead of going with a 5/8 .035 tube, try .75 x .028. Very similar weights, but one is a lot better in bending.

The highest stressed part of the car is generally the unsprung. It usually accounts for about 20% of your total weight. Adding one or two percent more to that number with increase your strengths exponentially. It is a very good return for investment. Over build your unsprung parts.

Don't forget, cones hit a-arms, peoples hands pick up the car by them, don't make them twigs.

[OspreysGoSWOOP]

Good points. I'll probably end up using 1.5 for braking/accel but I needed some calcs so I wasn't pulling numbers out of air.

Yea. I knew average accel wasn't a "robust" approach but our data is limited as a first year team and I needed to approximate something other than the numbers I see tossed out in the forums. Now I have a rough frame of reference.

Regarding the bending LCA. We'll be mounting the pushrod to our upright to avoid this problem.

Good point about the a-arms. I'll keep that in mind.

[Charlie]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by exFSAE:
2.5 FOS is pretty insane, dude. Even 1.2 is a really damn high FOS </div></BLOCKQUOTE>

It depends on the accuracy of your load cases. If you are making a very rough calculation as the poster is, a large FOS makes sense.

To say a FOS is 'insane' doesn't make any sense. I'd say the result is a fairly reasonable number.

If your 'really damn high' 1.2 FOS is used in this example, the part would be designed to 0.46 long G, which would likely be insufficient.

Now if the poster had a car with strain gauged parts and tested it, I would agree with a much smaller FOS on those numbers. But that isn't the case here.

[flavorPacket]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by OspreysGoSWOOP:
Regarding the bending LCA. We'll be mounting the pushrod to our upright to avoid this problem. </div></BLOCKQUOTE>

If you're having trouble calculating your total vehicle accelerations, you should not be putting pushrods on uprights. One of those things is fundamental, while the other is something that professionals still screw up even after doing sophisticated multibody dynamics analysis.

[oz_olly]

If you put the pushrod or pull rod on the upright wouldn't you have to make sure the centre of the spherical bearing coincides with the steering axis?

Surely it is much easier just to make sure the push rod force acts directly through the centre of the outer spherical on the control arm. Some amount of angular displacement will be seen through the full bump travel but the force component contributing to bending the control arm would be very small.

[Mike Cook]

In the rear you could mount your pushrod on the upright because it doesn't really steer, but this won't work for the front. I wouldn't recommend doing this on either axle though.

olly,

In my experience, usually you need to have your pushrod somewhere inboard of the lower ball joint for packaging reasons. But yes, ultimately you want your pushrod point as near to the lbj as possible.

[exFSAE]

You know what I meant, Charlie. Even for a team without dynamic load measurement, you can come up with some pretty firm upper bounds for load cases... eg when the car is gonna flip over. That's a pretty conservative approach as it is, and designing for 20% overload on top of that is a lot.

[flavorPacket]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by oz_olly:
If you put the pushrod or pull rod on the upright wouldn't you have to make sure the centre of the spherical bearing coincides with the steering axis? </div></BLOCKQUOTE>

typically you don't, actually. Thus the motion ratio changes with steer, and the steering feel changes with load, etc. It's more than picking a point and calling it OK.