Help me to understand

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rlaton
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Help me to understand

Post by rlaton »

I understand that the weight needs to go mostly to the back and where to put it and all that. I just don't understand WHY it being at the back helps. I know that everyone SAYS that it stores energy and keeps the car from slowing down on the flat. I still don't understand WHY. I also understand that the less weight you have on the front, the more sensitive alignment becomes and I understand why. Too little on the front and you get the "Dreaded Wiggles". But, assuming PERFECT alignment, why would 5 grams on the front, not be better than 25 grams on the front. So, here's my exact questions.....

1. WHY does moving the weight back make the car faster?

2. Does moving it back on a short wheel base effect the car the same way as a long wheel base?

3. If the track was a constant drop and had no flat, why would ANYTHING that you did with weight effect the speed of the car. Assuming that you don't move it too far back and cause the car to WIGGLE.

4. Lets say you dropped a 20 foot long TUBE from 100 feet up. The tube had 100 lbs of weight in the BOTTOM of it and the tube was exactly vertical. Would it hit the ground at the same exact instant that the same tube with the same 100 lbs of weight in the TOP instead of the BOTTOM would?
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Stan Pope
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Re: Help me to understand

Post by Stan Pope »

First visualize the path that the car's CM follows as the car's nose runs from start line to finish line. Do this three time: first for the CM just behind the front axles, then midway, and finally with the CM just ahead of the rear axles.

Compare the steepness of the three paths.

The steepest path is the one with the CM near the rear axles.

Now, if the lamp has not brightened, visualize a bit further...
three straight ramps connecting the car's CM location at the start line and the finish line. Which ramp is steepest?

Does the steepness of the ramp influence the time required for a car to traverse it?
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Re: Help me to understand

Post by rlaton »

Let me see if I understand. Is it simply because the weight doesn't reach the bottom of the hill for longer? When the weight reaches the bottom of the hill is it no longer falling. I guess when the weight is farther back, it pushes for longer? Is that the ENTIRE reason for your thinking?
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Stan Pope
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Re: Help me to understand

Post by Stan Pope »

Interpreting your question literally, neither yes nor no would avoid misleading you. Attempting to use either answer to develop your understanding and to build upon it, you would probably reach incorrect conclusions.

So, it is better that I suggest repeating the visualizations, and addressing the question regarding path steepness on time to traverse.

(BTW, the "weighted tube scenario" is not a sufficient analog to a PWD car, so it is useless to pursue it.)
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Re: Help me to understand

Post by pack529holycross »

rlaton wrote:Let me see if I understand. Is it simply because the weight doesn't reach the bottom of the hill for longer? When the weight reaches the bottom of the hill is it no longer falling. I guess when the weight is farther back, it pushes for longer? Is that the ENTIRE reason for your thinking?
Im not a rocket scientist, but the basics as I understand them are as follows:

The FARTHER a given mass falls on the incline, the MORE energy is translated to the vehicle on the flat part of the track ( A car that is 1/2 way up the hill won't travel as far as a car that is started at the top of the hill )

The more your CM is towards the BACK of the car, the greater the distance between its starting position ( at rest ) and the transition to the flat part of the track, providing the MOST energy possible for that translation.

How was that, Stan?

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Re: Help me to understand

Post by rlaton »

pack529holycross wrote:
rlaton wrote:Let me see if I understand. Is it simply because the weight doesn't reach the bottom of the hill for longer? When the weight reaches the bottom of the hill is it no longer falling. I guess when the weight is farther back, it pushes for longer? Is that the ENTIRE reason for your thinking?
Im not a rocket scientist, but the basics as I understand them are as follows:

The FARTHER a given mass falls on the incline, the MORE energy is translated to the vehicle on the flat part of the track ( A car that is 1/2 way up the hill won't travel as far as a car that is started at the top of the hill )

The more your CM is towards the BACK of the car, the greater the distance between its starting position ( at rest ) and the transition to the flat part of the track, providing the MOST energy possible for that translation.

How was that, Stan?

Nicholas


Okay, I think I got it. The weight is farther up the hill, therefore basically making the hill LONGER. Giving you MORE time to build up speed. So....

1. How does that help or hurt on a track that is a constant slope??

2. Alignment being excellent, what makes a short wheel base better or worse than a long wheelbase?
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Re: Help me to understand

Post by Stan Pope »

rlaton wrote:
pack529holycross wrote:
How was that, Stan?

Nicholas


Okay, I think I got it. The weight is farther up the hill, therefore basically making the hill LONGER.
Yes, on a normal track there is more hill and less flat.
rlaton wrote:Giving you MORE time to build up speed. So....
Yes, you might have more time, but analyzing by "time on the hill" can lead you off into the weeds. Think about the cars you have seen on a track. Ever see a rear-weighted car beat a mid-weighted car to the bottom of the slope. It happens a lot on a standard Cub Scout track profile!

Better to think in terms of the (CM) path and its slope (which controls how gravity affects the car) and the distance over which gravity can accelerate the car. "Times" depend upon how the car responds to its various environments.
rlaton wrote:1. How does that help or hurt on a track that is a constant slope??
If the track has constant slope, then he CM paths for all the CM locations have identical shape (length and slope.) So CM location does not confer an energy advantage.

CM location does affect the weight borne by the front and rear wheels. It may, therefore, be important to have just the right amount of weight on the front wheel to make good use of the rail to guide the car straight down the track.
rlaton wrote:2. Alignment being excellent, what makes a short wheel base better or worse than a long wheelbase?
On a track with curvature, the wheelbase affects the path that the CM follows. A short wheelbase car places the CM closer to the running surface during the curve. This actually makes the track a bit steeper earlier for the short wheelbase car!

A few builders may be able to prepare their car well enough to take advantage of that path difference.

Secondly, shortening the wheelbase affects the relationship between CM location and weight distribution. For a given distance between the rear axles and the CM, shortening the wheelbase increases the amount of weight carried by the front wheels. This, in turn plays with the toe-in needed for effective RR.
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Re: Help me to understand

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Stan Pope wrote:
rlaton wrote:1. How does that help or hurt on a track that is a constant slope??
If the track has constant slope, then he CM paths for all the CM locations have identical shape (length and slope.) So CM location does not confer an energy advantage.

CM location does affect the weight borne by the front and rear wheels. It may, therefore, be important to have just the right amount of weight on the front wheel to make good use of the rail to guide the car straight down the track.
Stan-Have you done any empirical optimization for a contant slope track? I hope to never run on one, but my curiosity has been piqued by a couple of your posts on this subject. Given my lack of real world experience, any pointers for my gedanken experiements?

Any particular recommendations for weight distribution of a constant slope track?
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Stan Pope
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Re: Help me to understand

Post by Stan Pope »

priority wrote:Stan-Have you done any empirical optimization for a contant slope track? I hope to never run on one, but my curiosity has been piqued by a couple of your posts on this subject. Given my lack of real world experience, any pointers for my gedanken experiements?

Any particular recommendations for weight distribution of a constant slope track?
Nope! Thankfully, those tracks are but legends in these parts! I think that I saw one in storage somewhere about 15 years ago, though.

Someone posted a "one off" comparison of Dead-On vs. RR on such a track recently, but RR has too many variables to be tossed overboard based on one sample. My guestimate is that it runs best with light toe and light front wheel weight, but that is all intuition. Next choice would be light toe and moderate front wheel weight.
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Re: Help me to understand

Post by GravityRacer »

<tosses match on fuel-soaked logs>

The amount of energy available (potential energy) to be converted to speed is related to the difference in height (start to finish) of the cg on the track:

PE=mgh.

Regardless of track configuration, the speed will be greater, all things considered, with a cg moved further back. There is all kinds of anecdotal evidence that is pointed to that disagrees with my assessment, but the cold hard physics is in the above formula.
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Stan Pope
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Re: Help me to understand

Post by Stan Pope »

GravityRacer wrote:<tosses match on fuel-soaked logs>

The amount of energy available (potential energy) to be converted to speed is related to the difference in height (start to finish) of the cg on the track:

PE=mgh.

Regardless of track configuration, the speed will be greater, all things considered, with a cg moved further back. There is all kinds of anecdotal evidence that is pointed to that disagrees with my assessment, but the cold hard physics is in the above formula.
Do the visualizations suggested early in this thread, but for a constant slope track. For each case, compute the difference in height between the car's CM when its nose is at the starting line and when its nose is at the finish line.

When you have correct input, then you can get correct output from the physics equations. Until then, GIGO.
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Re: Help me to understand

Post by gpraceman »

Stan Pope wrote:Thankfully, those tracks are but legends in these parts! I think that I saw one in storage somewhere about 15 years ago, though.
There is one track manufacturer that makes a constant slope track, Indy Products.

I don't know what their sales are like, but I have yet to see one of these tracks, even while surfing the web looking at race photos.
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Re: Help me to understand

Post by rlaton »

Stan, I still cannot VISUALIZE what you are asking me to visualize. Can you put it in diagram form? I don't see why the weight being "Further up the hill" makes any difference. The car is sitting on the track and the slope of the track does not change so I don't see how being higher up the hill makes any difference. One post suggested that if you take a car half way up a hill and release it, that it will not be as fast as the same car taken all the way to the top of the hill and released. I agree with that but don't see how it applies here. The distance from the release gate to the finish line doesn't change no matter how far back you put the weight. The car could be 30 feet long and the weight be on the back "Bumper" and it still only FALLS as far as the front of the car does to the finish line. I KNOW that you are right. I've seen what you are saying to be TRUE. I'm not disagreeing. I just cannot wrap my head around WHY. Please try again. Thank you in advance.
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Stan Pope
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Re: Help me to understand

Post by Stan Pope »

rlaton wrote: Stan, I still cannot VISUALIZE what you are asking me to visualize. Can you put it in diagram form? I don't see why the weight being "Further up the hill" makes any difference. The car is sitting on the track and the slope of the track does not change so I don't see how being higher up the hill makes any difference.
Too many questions mixed into the discussion involved has caused the answers to be combined with the wrong questions. Parse carefully, and it should become clearer.

On the constant slope track, the paths are totally equivalent and CM location does not affect the energy available. Any advantage from CM location must derive from some other cause, such as ability to use RR effectively.

On a traditional Cub Scout track, the slope is not constant and the energy differences are present and the CM paths are not equivalent.

Gotta run get ready to take Bride to the theater, now. If I don't lose the thought, I'll document another exercise that may be helpful tonight.
Stan
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Re: Help me to understand

Post by Darin McGrew »

There are images on the forum somewhere, but I couldn't find them. I'll do my best with ASCII art.

I'll diagram a typical track and a constant-slope track. For simplicity, I'll exaggerate the angles and distances. The difference will be the same with a typical derby car, only less extreme. I'll diagram three CM locations: front wheels, rear wheels, and half-way between the front and rear wheels.

Code: Select all

Typical Track                                       __
                                                   /  \
                                                  | () |___
                                                  /\__/    |
                                                 / /       |
                                                / /        |
                                               / /____     |
                                              / /     |    |
                                             / /      |    | 
                                          __/ /       |    |
                                         /  \/___     |    |
                                        | () |   |    |    |
                                         \__/    |    |    |
                                                 |    |    |
                                                 |    |    |
                                                 |    |    |
                                                 |    |    |
                                                 |    |    |
                                                 |    |    |
A = distance CM drops in front-weighted car      A    |    |
B = distance CM drops in mid-weighted car        |    B    |
C = distance CM drops in rear-weighted car       |    |    C
                                                 |    |    |
                                                 |    |    |
                                                 |    |    |
                                                 |    |    |
                                                 |    |    |
                                                 |    |    |
                                                 |    |    |
  __                   __                        |    |    |
 /  \_________________/  \                       |    |    |
| () |_______________| () |______________________|____|____|
 \__/                 \__/
Note that the CM drops further when it is further back on the car. The total distance traveled is the same in all three cases, because the starting pins and finish gate are a fixed distance from each other, and the nose of the car (and the entire car, for that matter) will always travel that distance. The further the CM drops, the more energy the car has and the faster it can travel on the level section.

Code: Select all

Constant-Slope Track                                __
                                                   /  \
                                                  | () |___
                                                  /\__/    |
                                                 / /       |
                                                / /        |
                                               / /____     |
                                              / /     |    |
                                             / /      |    | 
                                          __/ /       |    |
                                         /  \/___     |    |
                                        | () |   |    |    |
                                         \__/    |    |    |
                                                 |    |    |
                                                 |    |    |
                                                 |    |    |
A = distance CM drops in front-weighted car      A    |    |
B = distance CM drops in mid-weighted car        |    B    |
C = distance CM drops in rear-weighted car       |    |    C
                                                 |    |    |
            __                                   |    |    |
           /  \                                  |    |    |
          | () |_________________________________|____|____|
          /\__/                                  |    |
         / /                                     |    |
        / /                                      |    |
       / /_______________________________________|____|
      / /                                        |
     / /                                         |
  __/ /                                          |
 /  \/___________________________________________|
| () |
 \__/
Note that the CM drops the same distance, regardless of where it is located on the car. Again, the total distance traveled is the same in all three cases.

Note that in both cases, there may be advantages to placing the CM further back, independent of the energy issues illustrated above.
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