control arm load calculation

[titanium]

hi..i got a problem with the force resolution. i did the calculation but not sure whether i am on the right track

http://imageshack.us/photo/my-...es/859/95577745.jpg/ (http://imageshack.us/photo/my-images/859/95577745.jpg/)

? Fx = Fc + F1 cos? + F3 cos ? + F2 cos ø = 0
? Fy = W - F1 sin? + F3 sin ? - F2 sin ø = 0

Calculating moment about the roll centre (RC)-

? M = (Fc*e) – (W*c) – (F1 cos?*d) + (F1 sin?*a) – ( F3 cos ?*f ) – (F3 sin ? *b) – (F2 cos ø*f) + (F2 sin ø*b) =0

F1, F2, F3 can be found from the three equations by substituting the rest of the values from obtained geometry (in lotus suspension software).
i would really appreciate if somebody would Correct me if I am wrong some where

[JasperC]

That's simple statics so I guess what you're saying is correct (for a steady-state situation).

[Sormaz]

if you want to get really technical,
does ? Fx = 0 ?

[titanium]

Originally posted by JasperC:
That's simple statics so I guess what you're saying is correct (for a steady-state situation).

sir i was trying for dynamic condition so i added up the cornering force Fc too..

[titanium]

Originally posted by Sormaz:
if you want to get really technical,
does ? Fx = 0 ?

thanks for your reply sir. but during cornering shouldn't all the forces in the horizontal sum up to zero for avoiding vehicle roll about roll centre?

[Sormaz]

Originally posted by titanium:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Sormaz:
if you want to get really technical,
does ? Fx = 0 ?

thanks for your reply sir. but during cornering shouldn't all the forces in the horizontal sum up to zero for avoiding vehicle roll about roll centre? </div></BLOCKQUOTE>
Nope, that is handled in your third equation.
You should be able to get this

[Z]

Titanium,

What is the part you are having difficulty with?

Your image shows a VERY SIMPLE 2-D FBD of forces acting on a massless wheel assembly. I copied your sketch and used geometry (no equations) to get a rough idea of the force magnitudes in about ten seconds! (Pushrod in compression carrying most of Fc+W, wishbones in low tension.)

How did you get the magnitudes of Fc and W? (They are important for the rest of the problem.)

Why do you bother showing the RC? (It is irrelevant.)

What will you do when you have more complex dynamics problems to solve? (Your car moves in 3-D.)

Who is paying for your classes? (If it is your father I suggest he asks for his money back.)

Z

(PS. The rot really did set in when they stopped teaching Euclid. http://fsae.com/groupee_common/emoticons/icon_frown.gif )

[titanium]

considering the front quarter of the vehicle
W = (( total vehicle weight+ driver)*weight distribution)/2
(so that does not make the wheel massless)

the part i am having difficulty with-
i have seen most teams considering Fc to be 1.3g. how do they arrive at that value

RC has been shown cause i am considering vehicle roll about RC for calculating the moment.

[Z]

Titanium,

1. If your wheel has mass, then why have you not shown the gravitational and inertial forces passing through the wheel's CG?

2. Have you done a class called "Mechanics", (usually subdivided into Statics, Kinematics, and Dynamics)?

3. Has your school taught you Newton's Laws of Motion?

4. Or the Simple Law of Friction?

5. Or about Jean Le Rond d'Alembert, and his Principle?

6. Are you assuming that "RC" is the 2-D centre for kinematic motion of the car's body relative to ground?

7. If so, how have you come to this conclusion?

8. Have you heard of Euclid?

9. Are you paying for your education?

Please answer all above questions so that I can then help you.

Z

[BrandenC]

Originally posted by Z:
Titanium,

1. If your wheel has mass, then why have you not shown the gravitational and inertial forces passing through the wheel's CG?

2. Have you done a class called "Mechanics", (usually subdivided into Statics, Kinematics, and Dynamics)?

3. Has your school taught you Newton's Laws of Motion?

4. Or the Simple Law of Friction?

5. Or about Jean Le Rond d'Alembert, and his Principle?

6. Are you assuming that "RC" is the 2-D centre for kinematic motion of the car's body relative to ground?

7. If so, how have you come to this conclusion?

8. Have you heard of Euclid?

9. Are you paying for your education?

Please answer all above questions so that I can then help you.

Z

Aren't you being a bit harsh? Come on man, we've all got to start somewhere. Just cause you've got your head wrapped around vehicle kinematics doesn't mean everyone else will.

[BBolze]

well, it's just that he's making really simple mistakes at the very start. So they carry through and anything he calculates from this will be worthless.

I mean, he's assuming the mass of the vehicle in all one lump sum. ... This is suspension, the two masses are isolated from each-other through your spring/damper. You CAN'T consider the vehicle as one mass for this problem to do it properly. And because he's thinking as if he were in the car, he hasn't included the inertial force which would be what balances summation of Fx. The Kinematic RC is not whats important, He doesn't even have a CofG height????

This guy has paid tens of thousands of dollars for his education and he's not in turn invested himself.

[titanium]

Z,

though i was taught mechanics in my class, but reading the theory and applying it practically are two different things. being a beginner i tried my best but was still left with befogged ideas. i posted here to clear my concepts on practical application.


BBloze,

i still have few more questions.(sorry if you find them silly)

are the two masses you are referring to, sprung and unsprung mass?

does the addition of inertial mass include vehicle sprung mass+unsprung mass+ driver's weight or just the sprung mass+driver's weight.
(skipping the inertial mass was a big blunder on my part.dont know how i skipped it and didnt even realize)

why is the kinematic RC not important?

about what point should i take the moment?

[Z]

BrandenC,

As I have said before, I blame the education system.

It used to be that any junior draughtsman would solve this problem, easily! Sadly, these days it is not just Titanium, but also a great many "senior engineers" who struggle with these simple problems. Employing a mechanical, structural, civil, aero, etc., engineer who can not solve these problems effortlessly is a bit like employing an accountant who can not count past three! (And, yes, I reckon there are a lot of those about now too "Oh, well, the computer does that counting stuff for us now...".)

The crazy part about all this, is that it is really very easy. I had a similar discussion a few years ago, so I decided to experiment on my kids. After about ten minutes of "lecturing" my 12 year old son had grasped the basics. Encouraged, I started on my teenage daughter. There was a lot of "But Dad, I don't need to know this stuff...! (she had other exams coming up). Nevertheless, I persisted and half an hour later she could have solved Titanium's problem (maybe with a little prompting).

A lot of the utter codswallop that is promulgated by so called vehicle dynamics gurus these days is a result of this failed teaching system. All of the ADAMS templates, Lotus programs, HyperMegaRacerSims, etc., amount to nought if the user doesn't have a clue about the basics. My main concern is that these uneducated dimwits, having learnt to "talk the talk" regarding some CAE program, are now being allowed to design jumbo jets and nuclear reactors. Yet they have no basic understanding of what forces are, or how they work!

I stress again that vehicular "mechanics" is a very simple subject. It is not interplanetary rocket dynamics, and certainly not quantum mechanics. I reckon the older engineers working on the various space programs of the 1950s and 60s were the last generation that had a decent basic education (ie. one that included the Elements). They sent men to the moon using pencil, paper, and slide-rules (and, incidentally, manual machine tools). It comes as no surpise to me that any American wanting to go into space today has to hitch-hike with the Russians. And the Ruskies latest Mars probe is still stuck in Earth orbit! In fact, they've had no success in that direction for over twenty years... http://fsae.com/groupee_common/emoticons/icon_frown.gif
~~~o0o~~~

Anyhow, the only way that this situation will improve is if people start acknowledging that the current teaching system has failed. So, to prove my point http://fsae.com/groupee_common/emoticons/icon_smile.gif, here is a simple test;

1. Can someone, anyone, please post an accurate definition of what a "moment" is (as used by Titanium above)?

2. How can this "moment" help Titanium with his understanding of his car's performance?

Respondents scoring less than 100% should ask their school for a partial refund of fees. Those not confident enough to post any reply should ask for all their money back. http://fsae.com/groupee_common/emoticons/icon_smile.gif

Z

[BrandenC]

Z,

It appears that you have some very strong sentiments about this topic (current state of educational system). Maybe you should create a new thread dedicated to this topic to make it more visible.

I'll let someone else take a shot at your test (I may even ask for my money back http://fsae.com/groupee_common/emoticons/icon_smile.gif ).

Finally something for you to think about. You seem pretty quick to dismiss RC height, however it seems RC height will determine jacking forces that in turn will affect control arm loads, which is what we are after isn't it? Maybe RC height does have some importance.

I'll leave with this; I rarely visit these forums anymore but when I do (insert Dos Equis joke here http://fsae.com/groupee_common/emoticons/icon_biggrin.gif ) it pains me to see prominent veteran members being so quick to chastise students who are just learning the in's and out's of engineering and vehicle dynamics. This forum and FSAE in general are meant to help students studying engineering improve their ability to apply concepts from the classroom to a real world problem. While a professional engineer should be expected to understand how to perform simple force vector analysis, FSAE is not aimed at PE's it's aimed at students. As a student this forum was a goldmine for helping me get a handle on vehicle kinematics and dynamics, which have helped me immensely as an engineer. It would be nice to see this place continue to provide helpful feedback to those who come with questions, it (and FSAE as well) is truly is a great resource.

[LemonFool]

the part i am having difficulty with-
i have seen most teams considering Fc to be 1.3g. how do they arrive at that value

I would stop and understand the problem first. It seems like you have no idea what you're solving for. You need to find some sort of goal/problem, whether it's denoted by the rules or your performance. In this case, I would go looking at the dynamic events and a physics I topic called uniform circular motion.

After that makes sense, go draw yourself a picture and think about it. Maybe look into a bicycle model. I think you can find it in Gillespie's book. Better yet, digest Gillespie's book. Don't go thrashing on equations with your suspension geometry with no end. Re-read the constructive parts of this post, and then go back to your drawing. Things should (I hope) click a bit more. Until you can do that, the rest is garbage.

[Ben Coburn]

Re: "moment": this question seems so simple that it's setting off my trick question sensors, so I'm less confident than I probably should be, but:

He seems to be using it to mean "sum of the moments/torques from the applied loads", which is just one of the expressions that needs to be equal to zero for static equilibrium.

This isn't really useful for understanding performance as such, won't help you to make the car go faster, but it's sortof necessary for calculating A-arm loads, which is actually the topic in question.

The RC height, and indeed position in general, are quite irrelevant. If everything is at equilibrium the sum of the moments about any point at all is zero, there's nothing special about the RC for this problem.

This is the right general approach, and just solving those equations with the proper values for Fc and W will indeed give you the right answer. Getting reasonable values might be harder, although you can really just guess very high and be fine. I also recommend turning this into a computer program so that you don't have to do the algebra every time.

So, does Tech owe me a refund? :P

(misc things: I actually disagree with Z's analysis, I'd think that as drawn, and for the ~1.5 coefficient of friction I'd expect, that the lower a-arm will be in significant compression rather than light tension. I've also watched an entire FSAE team at Georgia Tech fail to reliably have a solution to this problem, so I feel his pain re: the education system)

edit because two posts in a row seems tacky:
BBolze: Insofar as this is just a free body diagram of the wheel/upright, you really can treat the car as just one lump mass. You also don't need to know anything at all about CG height. I guess it's possible Titanium is actually thinking of this as a FBD for the actual chassis, in which case maybe things are more complicated, but I still don't think CG height is relevant. I also don't think W/Fc would be in the drawing in that case.

[Z]

Unfortunately, I have timed this badly. I'm leaving civilisation for the next week, so no posts until back...

Nevertheless, I'd still like to hear anyone's ideas of what is a "moment". Preferably a detailed mathematical type definition together with its practical implications, rather than a simplistic "it's a rotational force". Also, what is the difference between "force", "couple", "torque", and "moment".

Meanwhile, here is something to think about;

https://lh3.googleusercontent.com/-vz_npv2BQ2c/Tv7174c1DnI/AAAAAAAAAIc/_6Iu2MwDSVU/s800/ControlArmForces.jpg

Not all working is shown, but no algebra is required for a good estimate of the forces. Just old school geometry. http://fsae.com/groupee_common/emoticons/icon_smile.gif

Z

[kcapitano]

Alright, I'm going to take a stab at this force, couple, moment, torque game. But, before I discredit myself, a few words on the educational system.

It is my firm belief that no university program can make you a good engineer, or whatever you are trying to become. University simply gives us the tools to become good engineers ourselves. Let's face it, we wouldn't be having this discussion had no one sat us down and told us about forces in the first place. IMO, anyone on these forums is, at the very least, taking steps to become a good engineer. So kudos to everyone!

Now for the definitions...I'm significantly less confident about this than I should be...

Force: A push or pull acting at point. Has both magnitude and direction.

Moment: The result of a force acting at some distance from a point. Calculated as the vector cross product of force and distance; therefore, has both magnitude and direction.

Couple: A pair of forces, equal in magnitude, opposite in direction and not colinear, acting on a body.

Torque: The resulting moment created by a system of forces. For a rigid body torque will be equal regardless of the point about which it is calculated.

Criticism cheerfully accepted.

[mech5496]

Originally posted by Z:

Meanwhile, here is something to think about;

https://lh3.googleusercontent.com/-vz_npv2BQ2c/Tv7174c1DnI/AAAAAAAAAIc/_6Iu2MwDSVU/s800/ControlArmForces.jpg

Not all working is shown, but no algebra is required for a good estimate of the forces. Just old school geometry. http://fsae.com/groupee_common/emoticons/icon_smile.gif

Z

Or as we call it in Greece, "vector analysis". Actually I was taught that particular class in Junior Highschool sometime (confirmed; I just asked my sister-aged 13-to solve a simple 2-vectors sum problem using geometry and she did it in like 3 minutes)... Back on the problem, I dare to disagree with you Z. On the above picture, you can calculate ALMOST everything based on vector analysis, except that you do not know the magnitude of F1 and F3. They could be as drawned; or F3 could have the opposite direction and same magnitude-in that case, F1 would have the same direction but a magnitude of F1'=|F1+2F3|....Actually they could have any combination of magnitudes as long as their vector sum remains the same (F1+F3 vector shown above); think about it! You need an extra equation... http://fsae.com/groupee_common/emoticons/icon_wink.gif

[Zac]

Originally posted by titanium:
Z,

though i was taught mechanics in my class, but reading the theory and applying it practically are two different things. being a beginner i tried my best but was still left with befogged ideas. i posted here to clear my concepts on practical application.


BBloze,

i still have few more questions.(sorry if you find them silly)

are the two masses you are referring to, sprung and unsprung mass?

does the addition of inertial mass include vehicle sprung mass+unsprung mass+ driver's weight or just the sprung mass+driver's weight.
(skipping the inertial mass was a big blunder on my part.dont know how i skipped it and didnt even realize)

why is the kinematic RC not important?

about what point should i take the moment?

sprung mass is the mass supported by the vehicle suspension (ie. driver, body, cargo, engine, etc.). unsprung mass is the mass that is not supported by the suspension (tire, wheel, brakes). The weight/inertia of the suspension links themselves are a gray area. For simpler analysis they are often ignored, for more complex analysis (ie. multibody simulation) they are somewhat important.

the crux of your problem (and likely what has inflamed Z) is that you really don't know (or at least say) what the goal of your analysis is and/or understand the assumptions you are making.

First, the sum of forces (or moments) does not equal zero in a dynamic system. F = m*a. Racecars sure like to accelerate and do have some mass (despite our best efforts) so you will have to include inertial terms in a dynamic analysis (do you need to do a dynamic analysis?).

Second, you are correct that in the front view, you will need to have three equations in order to solve for your forces (three unknowns). Summing lateral force and vertical force is pretty straight forward trig (or vector analysis if you're into graphical techniques). You can sum those moments about the RC, or any other arbitrary point but that only makes it more difficult for you. You might as well use a point that limits the number of forces to solve for, like the outer pin joint of the lower control arm. Three equations, three unknowns, simple algebra at this point.

Third, drawing a FBD is a good start. But think about what you want to draw a free body diagram of. If you're trying to figure out the forces being reacted through your suspension links a FBD of just the unsprung mass will be a good start.

Fourth, W and Fc aren't really knowns are they? Up until this point all we've really done is generate a function that spits out the forces in your suspension linkages given more or less arbitrary values for W and Fc (along with ride height and body roll angle). Fc being equal to 1.3*W or whatever arbitrary level of friction you decide to solve for is fine, but will the car have equal load on the inside and outside wheels? Assuming a CoG above the ground plane, simple mechanics says no. You might want to look at chapter 18 of Race Car Vehicle Dynamics (or remember your lessons on summing forces and moments and read up on compatibility equations or Castigliano's therom). Much like a 4-legged coffee table a race car is under-defined from a statics standpoint.

[titanium]

Z

http://img408.imageshack.us/img408/2870/forces.png


obtaining the angles by geometry and applying sine law we get

[(Fc+W)/sin?] = [(F1+F3)/sin?] = [F2/sin?]
we can find F2 but F1 and F3 still remains unknown.



Zac


Third, drawing a FBD is a good start. But think about what you want to draw a free body diagram of. If you're trying to figure out the forces being reacted through your suspension links a FBD of just the unsprung mass will be a good start.

for calculating the forces through the suspension links doesn't the sprung mass also come into picture?

shouldn't W be equal to sprung+unsprung mass?

[Edward M. Kasprzak]

The vector sum of F1 and F3 must equal the resultant F1+F3 vector so this, too, comes out of the geometric solution. The directions of all three vectors are known, along with the length of F1+F3.

Now ask yourself this: How do you interpret the diagram when F1 and F3 are parallel? Do you need more information, or can the unique solution for F1 and F3 still be obtained from information in the resulting sketch?

To me, the "vector analysis" and the "geometric solution" go hand in hand. I do the math and plot the resulting vectors. Being able to do a hand check is always useful for verifying that the results make sense.

As for what goes into W, this is a modeling decision. It depends on how accurate you want/need your results to be, the range of conditions across which the results will be valid and how complicated a model you're willing to work with. Whatever you do, define your assumptions carefully and think about how they affect your results.

[Zac]

Originally posted by titanium:

Zac

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Third, drawing a FBD is a good start. But think about what you want to draw a free body diagram of. If you're trying to figure out the forces being reacted through your suspension links a FBD of just the unsprung mass will be a good start.

for calculating the forces through the suspension links doesn't the sprung mass also come into picture?

shouldn't W be equal to sprung+unsprung mass? </div></BLOCKQUOTE>

It is, but not in the way you're thinking. A FSAE vehicle has 4 wheels. All 4 wheels are carrying a portion of the vehicle's mass (though sometimes 1 or 2 wheels lift...). Unless very specific conditions are met, all the wheels carry a different amount of load. Cornering near the limit is not such a condition.

[swong46]

Originally posted by Zac:Cornering near the limit is not such a condition.

What is the defn. of limit here? He is calculating for A-Arm loads, not vehicle dynamics; so why not calculate at the limit?

My defn. of limit is the guy who you designed this car for drops if off his truck landing on one wheel sending that one damper to its bump stop! Or the car hits an imaginary large bump and the stiff/high freq. nature of these cars causes all the weight to be on one wheel.

[Ben Coburn]

Both of those will result in a somewhat more complicated load case/calculation than is shown here.

That said, the basic idea seems good: Just assume full weight transfer to the outside wheels, make a relatively high guess at your cornering acceleration/coefficient of friction, and call it a day. The difference between this and a more complicated/accurate calculation is probably small enough that you don't even get to go down a tube size, so spend your time somewhere else.

[Zac]

Originally posted by swong46:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Zac:Cornering near the limit is not such a condition.

What is the defn. of limit here? He is calculating for A-Arm loads, not vehicle dynamics; so why not calculate at the limit?

My defn. of limit is the guy who you designed this car for drops if off his truck landing on one wheel sending that one damper to its bump stop! Or the car hits an imaginary large bump and the stiff/high freq. nature of these cars causes all the weight to be on one wheel. </div></BLOCKQUOTE>

If you want to use something like that as your load case that's fine. All I was trying to get at is that under cornering the normal load acting on the tire will be something other than the nominal corner weight.

[Z]

Back from enforced leisure period, so...
~~~o0o~~~

Kcapitano,

You say;
-------------------------------------------------------------------
"Alright, I'm going to take a stab at this force, couple, moment, torque game. ...

Force: A push or pull acting at point. Has both magnitude and direction.

Moment: The result of a force acting at some distance from a point. Calculated as the vector cross product of force and distance; therefore, has both magnitude and direction.

Couple: A pair of forces, equal in magnitude, opposite in direction and not colinear, acting on a body. [and???]

Torque: The resulting moment created by a system of forces. For a rigid body torque will be equal regardless of the point about which it is calculated.

Criticism cheerfully accepted."
--------------------------------------------------------------------

The above is pretty much what is taught these days, but far too many mistakes for my liking (mistakes emphasised above).

I suggest you ask for a partial refund of fees. http://fsae.com/groupee_common/emoticons/icon_smile.gif Or at least a free and long clarification session with your Mechanics teacher.

More below, but briefly here;

Force: In Mechanics a force is considered to do its thing along a "Line of Action". No ONE point on the LoA is special. ANY point on the LoA is as good as any other. It is very important to remember that, in Mechanics, a force is a "sliding vector". (Structural analysis is different. It is a field of Engineering, while Mechanics is Applied Mathematics.)

Moment: What sort of "result"? The "vector cross product" part is correct, but does the resulting vector have an interpretation in the real world, in the same way that the mathematical abstraction of a linear "force vector" might correspond to a push or pull? That is, is a "moment" like a "purely rotational force"? Is it the same as a "couple", or a "torque"? Personally, I don't think so (see below). Unfortunately, many "experts" claim that moments, couples, and torques are all pretty much the same (see Glossary (page 491) of Blundell and Harty's "Multibody ... Vehicle Dynamics" for typically nonsensical definitions, together with other classical clangers).

Couple: Usually depicted as a single vector with magnitude and direction (so more than just "a pair of forces"). Importantly, unlike a force, a couple is a "free vector".

Torque: IF defined as a "resulting moment", then it is most definitely NOT "equal regardless of the point".
~~~o0o~~~

Harry and Titanium,

So, you both claim that F1 and F3 are not uniquely determined in my previous sketch?

Tsk, tsk, tsk...

Mark = F! http://fsae.com/groupee_common/emoticons/icon_frown.gif

Since it is such a simple problem, I suggest asking for a full refund! http://fsae.com/groupee_common/emoticons/icon_smile.gif
~~~o0o~~~

Edward,

You say;
------------------------------------------------------------------------------------
"Now ask yourself this: How do you interpret the diagram when F1 and F3 are parallel? Do you need more information, or can the unique solution for F1 and F3 still be obtained from information in the resulting sketch?"
------------------------------------------------------------------------------------

The solution is simple, and you don't even have to go off to infinity where parallel lines intersect. And you definitely do NOT need to use moments. The method is in all the old textbooks...
~~~~~~~~o0o~~~~~~~~

[Rant Mode=Full On]

I repeat that these types of problems used to be routinely solved by junior draughtsman and machinists. This stuff was taught in trade schools to the, err, "less bright" young boys. It was a prerequisite for ENTERING an engineering degree. Real "graduated engineers" were expected to do this stuff blindfolded. In a dark room... While asleep.....

Lately, IMO, the education system has spent far too much time focusing on the latest newfangled gimmickry - "Listen carefully students. You've got to perform this complex sequence of algebraic manipulations (or, more recently, "sequence of key strokes"), or else the system crashes...". And what's the result? Well, I now see several generations of engineers who have barely the foggiest idea about those most basic of engineering elements, namely forces and motions.

What next? I really fear it won't be long before professional engineers can't count past three. "Oh, our computers do all that menial stuff for us now. We focus on the more important business, of, well, business...". And shortly thereafter we all slide into the next Dark Ages. (Well, it might take a couple of hundred years, but nothing to look forward to...)

I doubt my efforts here are going to be of much use in stopping this decline. Nevertheless, for my own peace of mind (so I can say "I tried"), here are some miscellaneous mechanical thoughts.

1. Nowadays, due to the lack of any foundational examples of clear thinking (like the Elements), almost nobody bothers to define anything. So if you find yourself reading any technical matter that doesn't start with an attempt at clear definitions, then I suggest you stop reading. (BTW, this is a rant, so I'm not giving definitions here. http://fsae.com/groupee_common/emoticons/icon_smile.gif)

2. Some history. After western society spent 500 years climbing out of the last 500 year long Dark Ages (by studying Euclid), Stevin and others developed the geometric methods of Statics in the late 1500s. In the middle 1800s Hamilton developed Quaternion theory (see below). It was only in the late 1800s, with the motivation of understanding Electro-Magnetics, that Vector theory was fully developed.

3. Today Vector theory is treated as a "Swiss army knife" solution to a whole lot of technical problems. Indeed, it works really well for many problems, but is also hopeless for some others. Unfortunately, it seems that the vanity of the "education system" as a whole precludes it from admitting to the shortcomings of Vector theory. It is easier to ignore the flaws, which is not good education.

4. Modern Vector theory is a field of Applied Mathematics, so all its vectors and operators are used in pretty much the same way. But when applying these maths tools to real world problems we have to interpret the abstract vectors and their corresponding real phenomena in different ways. So we have to make the important distinction between real phenomena that behave as either "fixed", "free", or "sliding" vectors.

5. Here are some examples:
a) Using the Vector (Cross) product, we define a "Moment" as M=FxR.
b) Using the Vector (Cross) product, we define a "Velocity" as V=WxR.
In both these examples we might see M and V as fixed vectors (dependent on their points (the R's)), and F and W as sliding vectors (same effect anywhere along LoA).
Looks really neat, eh? One all-encompassing theory that covers BOTH forces AND motions! Wow!

6. Except that the rotational moment M corresponds to the linear velocity V, and the linear force F corresponds to the rotational velocity W. Huh??? It gets worse. If, rather than considering rotational velocity W (Omega), we instead consider rotational displacement Theta, then no more vectors! Nope, they don't work. http://fsae.com/groupee_common/emoticons/icon_frown.gif We instead have to use Hamilton's older Quaternion theory. So vectors are good for linear and rotational forces, velocities, and accelerations, and also for linear displacements, but NOT for rotational displacements. Bummer!

7. When considering the resultant of an arbitrary 3-D force system it is common to specify the "dual vector quantity" F+T at a certain point (with T = a couple, NOT a moment). With 3-D motions the "vector dual" is W+V. These vector duals lead to the concepts of a "Force Screw" (aka a "wrench"), and a "Motion Screw" (aka an "instantaneous screw axis" or "ISA"). The fundamental nature of these two "Screws", together with their usefulness, and frankly their simplicity, means all engineering students should be taught these during their first lesson of Mechanics 1.01. Incredibly, very few practising engineers, especially in the auto industry, have even heard of them!

8. To sum up, the "one tool does everything" approach to using modern Vector theory doesn't work well unless its limitations are made clear. Nowadays this doesn't happen. Perhaps the best way to do this in Statics is by teaching the old fashioned, and very simple, geometric approach first (a bit like learning to use a simple one-piece knife before trying to open up a Swiss army knife). There are only two really important rules to remember: 1) Forces can slide anywhere along their LoAs (but must not move sideways!). 2) Forces can be combined or decomposed using the "Parallelogram Method". The fact that a couple is a free vector is provable from these two axioms.

[Rant Mode=Off]
~~~o0o~~~

But while I'm here...

Zac says;
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"... the sum of forces (or moments) does not equal zero in a dynamic system. F = m*a. Racecars sure like to accelerate and do have some mass (despite our best efforts) so you will have to include inertial terms in a dynamic analysis ..."
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This suggests that "inertial forces" are not like other "real forces" (which is current teaching). That is, inertia has to be treated differently to the four real forces of Strong-Nuclear, Weak-Nuclear, EM, and Gravity. Note that Mechanics primarily deals with only EM (= inter-atomic push/pulls) and Gravity forces.

Has anyone heard a reasonable argument to support this claim?

The only arguments supporting "inertia is NOT a real force" that I have ever heard are, in essence, exactly the same as those that support the notion that certain other physical phenomena (eg. EM, gravity) ARE "real forces". The craziest part of this spectacular double standard is that almost nobody questions it! The education system is truly going down the plug-hole.
~~~o0o~~~

One last one...

Titanium,

If your calculations are for the purpose of structurally sizing your control arms, then you are heading for some embarrassing failures.

The static corner weight W (=~1/4 x total-weight), plus the slightly larger Fc, is nowhere near as large as the loads the arms will realistically see. I suggest calculating control-arm/rod-end sizes based on a pair of compressive forces of at least 3 x total-car-weight, acting on LoA between wheelprint and car CG. This is roughly equivalent to the car "dropped on one corner" from a very low truck (ie. dropped less than 0.3m, for ~3g acceleration).

Z

[Edward M. Kasprzak]

Edward,

You say;
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"Now ask yourself this: How do you interpret the diagram when F1 and F3 are parallel? Do you need more information, or can the unique solution for F1 and F3 still be obtained from information in the resulting sketch?"
------------------------------------------------------------------------------------

The solution is simple, and you don't even have to go off to infinity where parallel lines intersect.

Yes, it is simple. If you're new to suspension loads & analysis this is a good case to think about. In my opinion this is one of those questions that promotes understanding (or, said in a slightly different way, a "feel" for what's happening)--beyond the ability to calculate a solution.

[Zac]

This suggests that "inertial forces" are not like other "real forces" (which is current teaching). That is, inertia has to be treated differently to the four real forces of Strong-Nuclear, Weak-Nuclear, EM, and Gravity. Note that Mechanics primarily deals with only EM (= inter-atomic push/pulls) and Gravity forces.

Has anyone heard a reasonable argument to support this claim?

The only arguments supporting "inertia is NOT a real force" that I have ever heard are, in essence, exactly the same as those that support the notion that certain other physical phenomena (eg. EM, gravity) ARE "real forces". The craziest part of this spectacular double standard is that almost nobody questions it! The education system is truly going down the plug-hole.

A couple points I'd like to make. First, forces cause accelerations, it isn't the other way around. Second, all I was saying is that if A = 0, the inertial terms become zero. I don't particularly care if inertia is a real force or a fake force, or a red force or a blue force.

[Z]

Zac,

I wasn't saying that you are at fault.

Again, I was pointing out that the education system is propagating a very confusing double standard. Specifically, there are countless teachers, textbooks, websites, etc., that say;

"YOU MUST NOT THINK OF INERTIA AS BEING A REAL FORCE."

Your "don't care" comment above suggests that you are not actually sure yourself??? http://fsae.com/groupee_common/emoticons/icon_smile.gif

It is not unusual to be unsure. Many vehicle dynamics textbooks have figures with lots of force arrows labelled "Fx1", "Fy2", etc., and then the odd-one-out arrow labelled simply "ma". Why no "F"? Is it a force, or, err, acceleration???

My point is, why not just call the quite obviously "inertial" force "Finertia", or something similar? The answer is "peer group pressure". This fear of calling a spade a spade (a sort of "political correctness") is one of the biggest sources of confusion for students of mechanics.
~~~o0o~~~

So, for what it is worth;

!!!!!!!! INERTIAL FORCES ARE REAL FORCES !!!!!!!!!

Yes, you can treat them just like any other force!

So, if a massive particle is being accelerated rightward by a force Fx, then it also has a leftward "inertial force" Fi acting on it. To help understand this you can draw a simple vector diagram with all the forces (Fx & Fi) summing to zero.

In the case of a rigid body Fi acts along a LoA that passes through the CG, and there may also be an inertial couple Ti acting on the body.
~~~o0o~~~

It is interesting to note that when a car is stationary it has the acceptably real gravitational force Fg acting on it (this proportional to the relative position between car and Earth).

When the car moves with velocity V, it has the acceptably real aerodynamic force Faero acting on it (this roughly proportional to relative velocity-squared between car and air).

When the car accelerates at A, it seems to have an inertial force Fi acting on it (this proportional to relative acceleration between car and absolute space), but it is NOT ACCEPTABLE to call this phenomena a REAL force!

For those who think this is acceptable, please go to your school's Physics department and tell them that. (I suggest you give yourselves plenty of time... http://fsae.com/groupee_common/emoticons/icon_biggrin.gif)

Z

[DougMilliken]

Originally posted by Z:
Again, I was pointing out that the education system is propagating a very confusing double standard.
Z, is this what you are trying to say?
http://xkcd.com/123/

[Z]

Doug,

Yes, that's one of the few that suggest that inertial (eg. centrifugal) forces ARE "real". There are a lot more sites that insist the opposite, namely NOT REAL (in fact, they shout it a lot louder than that http://fsae.com/groupee_common/emoticons/icon_smile.gif).

Funnily enough, there was a "Big Bang Theory" re-run on just last night that had one of the geeks explaining how "it's not centrifugal, but centripetal force...".

For anyone interested in this issue, it boils down to Newton's concept of absolute space, versus Mach's ideas of relativity (later taken up by Einstein). There are a few papers available on the web explaining inertia as a low strength component of gravity, but proportional to 1/R, so dominated by the large amount of matter at large distances, and thus quite strong. But these versions, though going back ~200 years, are still in the minority, and in polite scientific company we should never mention the elephant in the room. http://fsae.com/groupee_common/emoticons/icon_smile.gif

The biggest problem with this whole issue, IMO, is that the scientific establishment, and hence also the education system, presents itself as having (almost) all the answers. "Inertia? Oh, we solved that problem ages ago...". A bit more honesty about how little is truly understood would be better for all.

Eg. Try googling "Twins Paradox". The majority of sites claim this is resolved by Special Relativity, hence no problem. A minority claim (rightly) that the SR versions are flawed and General Relativity is required for a resolution. A minority of these claim that even with GR the paradox remains!

Z