COM....How Clustered? How High or Low?

[Stan Pope]

I understand that the CM has a lower average velocity when it is raised, but doesn't the "low" nose of that same car have a higher velocity than the raised CM? after all, it's the nose that trips the finish sensor not the CM.

True, but it is the CM that the physics describe. The nose of the car has a fixed relationship to the CM in your car. So, you could think about it as if the finish line for this car triggering when the CM passes a point a few inches prior to the finish line. You are unnecessarily complicating your thinking.

Don't give up on "understanding." For most of us, it takes a lot of work.

Go forth and "break a law of physics" ... if you can! When we find differences between what "we know" and what "actually is", then real learning begins!

[M7 Racing]

Reading through these post made be think about stacking our weights for future cars. Is there any downside to puting a row of weight right over the rear axel or do you always want to put it in front or in back of the rear axel?

[Teeeman]

It's all about the weight distribution...

the Center of Mass (center of gravity) is going to be a cumulative function of where the mass of the car is distributed...

the simple check... balance the car on a ruler's edge (or something even sharper)...

the balance point (front to rear) will be under the CG aligned with gravity.

if you want the balance point to shift rearward, adding weight behind that point will shift the balance point in the direction of the weight...

ditto for forward...

the further you place the weight from the balance point, the more it will move the "new" balance point.

There is probably a Java applet already on the web that would work great, I'll look for it...

or I can give the math to someone if they can program Java...

-Terry

[joe]

Yes, usually you have to put some behind the rear axle to be able to lower the weight on the front end enough. Ideally you would want it over the axle and towards the centerline if possible. I know we've gone 'round and 'round on this, and if I get into a fight with a physicist I'll be unarmed. But before you stack your weights very high, note how much time you can gain with this configuration under PERFECT conditions. I think Michael Lastufka said 1/2" if your Center of Mass was 2 1/2 - 3 inches above the track? This is with perfect wheels, perfect track, and -- if I recall correctly -- no air resistance. Don't take my word for it, check his website, Stan posted it somewhere in here recently. No, I can't prove it, but I think you'll be ahead if you keep your CM lower. Think risk vs. reward.

[dknowles67]

Simply put, as a PWD car rolls down the track, it rotates about the center of mass (starts angled down, and rotates until it is level).
The COM is almost always between the axles, or the car will be unstable (pop wheelies)
The further away from the center of mass the weight is, the more energy is consumed rotating it about the center of mass.
With the weight high, (the back starts higher than the front), you have more PE to start with (PE = mgh (m=mass constant 5 oz, g=constant 9.8m/s/s h=height).
But you will consume more of it, as you have to rotate all that mass about the center. That's one reason why, generally speaking, thin flat cars with the weight concentrated between the axles go faster.
Track conditions will also play an important factor.
A bumpy crooked track will cause the car to bounce on the way down, causing more rotation about the center of mass.
A perfectly smooth track will cause less of this.
Weight concentrated around the center of mass helps more on a bumpy track.

[Stan Pope]

With the weight high, (the back starts higher than the front), you have more PE to start with (PE = mgh (m=mass constant 5 oz, g=constant 9.8m/s/s h=height).

Recheck this assertion ... The important aspect is the change in height as car goes from start line to finish line. Increasing CM away from axle plane reduces the change in height in the formula!

[Mike Doyle]

Here's my take (for what's worth):

Track type and conditions will play a major role in your selection of optimal weight placement, there is no *one* correct answer for every situation.

Secondly, while the shorter path for the CM results in slightly quicker run times, it requires radical placement that offsets the advantage it presents. If you factor in anything but a perfect track set-up, then it figures even more negatively.

With regards to weight placement, our tact has always been lower with an aggressive CG placement. Lower weight results in greater stability allowing for a more aggressive rearward weight placement equating to a higher PE on the initial slope.

As with anything in life, moderation is the key, too aggressive placement of the CG will result in a wandering front end that will quickly drain any gains from your speed account

If you're the type that benefits from visual aids, imagine a pick-up truck with refridgerators in the back, now imagine their effect on stability if they're stacked vertically -vs- laying down side by side.

[dknowles67]

Recheck this assertion ...

Correct, the change in height is where the Potential Energy comes from.
On a standard track, the back of the car starts higher than the front, and finishes level. Therefore, the back of the car changes height more than the front. Therefore more weight in the back will create more Potential Energy.

Is that what you were getting at?

Or were you trying to point out that putting weight on top of the car (high) as opposed to the bottom of the car doesn't change the PE much, since the weight falls the same distance?

Or was it something more complicated involving the geometry (trigonometry) of the car sitting on an incline, and the effects of different weight positions on the center of mass?
(Imagine a 3" antenna on top of the car (in the center) with a 3oz weight at the end - weight would be high on the level track, but depending on the slope of the track at the starting point, the weight may be below the back of the car)

Sorry, I didn't mean to imply something that wasn't true, but I was trying to briefly explain that it requires more energy to rotate mass, the further it is from the center of rotation. I got in too big a hurry to make the point.

Now I'm thinking about what I said before.
Does the mass rotate about the center of gravity?
It seems like the mass rotates about the front axle.
I think for physics equations, you can assume the mass rotates about the center of gravity.
That's why concentrating the mass at a single point is better.
I remember reading about this on the forum somewhere.

[Den_Leader]

...Or were you trying to point out that putting weight on top of the car (high) as opposed to the bottom of the car doesn't change the PE much, since the weight falls the same distance?...

In thisthread, Terry posted the diagram below for illustration. It graphically displays the difference in PE for a high and low CM. PE is greater for a low CM because a low CM actually falls further, not the same distance.

[dknowles67]

Ahh, the complicated Trigonometry I was referring to.
Notice in the diagram, the low CM is much closer to the back of the car, as it sits on the incline (not necessarily the back when level).

Assuming the CM was the same distance front to back (as the car sits on the incline), the change in height would be much less. That is, if the line between the low CM, and high CM were perpendicular to level, instead of perpindicular to the track.

But I can see where my original statement was misleading.

Cm High must be taken to mean Cm where it falls the greatest distance, not where it is in relation to the wheels on a level surface.

Boy, you guys sure know how to keep someone on their toes.

[Den_Leader]

Notice in the diagram, the low CM is much closer to the back of the car, as it sits on the incline (not necessarily the back when level).

In the diagram, the distance of the CM from the back of the car does not change for the high or low version.

Assuming the CM was the same distance front to back (as the car sits on the incline), the change in height would be much less. That is, if the line between the low CM, and high CM were perpendicular to level, instead of perpindicular to the track.

The CM starts at an incline and ends at level for both high and low. The important aspect is the length of the distance between the dots on lines representing the PE. You need to compare the distance between HighCM on ramp and HighCM on flat to the distance between LowCM on ramp and LowCM on flat.

Cm High must be taken to mean Cm where it falls the greatest distance, not where it is in relation to the wheels on a level surface.

Actually, no, CM High must be taken to mean where it is in relation to the wheels on a level surface. Stick with it you'll see the light. Please don't mistake me for someone who knows it all. I sure ain't that person. I screw up daily.

[dknowles67]

Not to argue too much, but you missed my whole point.
I think I understand the concept.

I can see how talking about front/back of the car with a reference to the car on a level surface will make it easier for everyone to understand, sorry I deviated from that, and caused confusion.

In my original post, I was referring to the front/back of the car as it sits on the incline. I didn't mean everyone has to change their way of thinking.
Only that to understand my original post, you would have to interpret the meaning the way I did when I posted it.

This is all moving away the the point I was originally making about losing energy as the mass rotates if the weight is far from the center.

I thought I was wrong once, but I was mistaken .

[Den_Leader]

... I thought I was wrong once, but I was mistaken .

I love it!

[Darin McGrew]

Does the mass rotate about the center of gravity?

It's easiest to evaluate the energy required for rotating the car if you assume that it does. You can do it with some other point of reference (e.g., the axles, or where the wheels contact the track), but the equations become more complicated, harder to analyze, and harder to explain without actually working the math.

Keep in mind that a derby car is a complex system. It's easier to isolate each effect, evaluating them separately, rather than evaluating them all at once. When we evaluate something else (e.g., friction, or how high/low to position the weight above the track), then we may choose a different frame of reference.

So when we evaluate how clustered around the COM the car's mass should be, all we look at is the energy required to rotate the car about its COM. Every time the car rotates around its COM, it looses energy. But how much energy?

For our purposes, the energy required to rotate an object depends on three things: the mass, the lever arm (the distance from the center of rotation), and how it rotates (speed, abruptness, etc.). Since the total mass is fixed at 5 oz, and we can't control how the car rotates (a function of the shape and smoothness of the track), the way to reduce the energy required to rotate the car is to reduce the distance from the center of rotation. That's why more compact weight has a (slight) advantage over more distributed weight.

[Stan Pope]

Simply put, as a PWD car rolls down the track, it rotates about the center of mass (starts angled down, and rotates until it is level).

Backing off all the digressions... This depiction does not really help understanding the losses involved in rotation.

If the car were thrown, any rotation in the car would be about the CM. But since throwing is not the useful method of movement, lets ignore it.
On a track, the car is constrained by the track, either its plan in the case of the curvature or its errors in the case of its roughness.

Looking first at the curvature, the center of the car's rotation is the center of the track's curvature at that point along the track. Raising the CM in the car decreases the radius of the CM motion, and that reduces the energy used (which was already very small anyway.) Most of the energy exchange is PE to linear KE. A very small amount goes into rotational KE. And the rate of transfer to rotational KE is very slow as compared to impacts.

Next looking at impediments ... bumps along the track.
A bump that affects both front wheels equally results in body rotation about the rear axles... which a fraction of a second later, when the rear wheels hit the bump, a body rotation about the front axles.

A bump that affect the wheels on one side of the car creates a very different rotation ... probably around a line through the outer bottom edge of the wheel bores on the opposite side. If the bump is big enough, the rotation continues around a line through the outer bottom edge of the wheel treads on the opposite side.

Impacts cause a sudden movement in the car's mass. The rate of that movement affects the energy required to cause the movement. Also the distance affects the energy required. The larger the radius, the more distance and energy.

IIRC, the square of the radius is involved in energy computations, which could explain the value of centralized CM on tracks with rough section transitions. Maybe a real engineer could shed some light here!