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weight
Feb 13, 2008 6:41:23 GMT -7
Post by Fat Boy on Feb 13, 2008 6:41:23 GMT -7
any comments on this weight triangulation theory ?
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weight
Feb 13, 2008 7:24:59 GMT -7
Post by docb on Feb 13, 2008 7:24:59 GMT -7
Huh?
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weight
Feb 13, 2008 11:49:52 GMT -7
Post by Fat Boy on Feb 13, 2008 11:49:52 GMT -7
theory being if you run a raised wheel , you would triangulate the weight on the dominate side as to have more on the side of your dominate wheel i think i said that right
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weight
Feb 13, 2008 15:45:33 GMT -7
Post by A-Line Performance on Feb 13, 2008 15:45:33 GMT -7
If most of your weight is between the rear wheels and very little is on the front or dominant, It would take a lot of shift to make any difference. I have not tried moving the weight to one side.
Of the cars I have seen with that kind of weight placement, none have been better than the cars with centered weight - ie: those owned by 2Fast4U as an example. His cars are still the ones to beat in the Nation.
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badger
PDDR Forum Member
[Mo0:0]
Posts: 2
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weight
May 19, 2009 20:46:11 GMT -7
Post by badger on May 19, 2009 20:46:11 GMT -7
Is this the same thing as positioning the weight so that the vector force on both the rear wheels is the same? (presumably moving the weight left or right toward the dominant wheel?) Seems like the friction would be higher on the non-dominant wheel if you DIDN'T do this...
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weight
May 19, 2009 22:50:55 GMT -7
Post by W Racing on May 19, 2009 22:50:55 GMT -7
Badger, That is exactly the theory. I have done the math to calculate what it would take. My thoughts that keep me from trying it: 1) Too many record breaking cars do not use it... When I look at sunburst for example I see the tungsten disc planted right in the middle. 2) Shifting weight to one side makes you stack it higher. Loosing potential energy. 3) Less weight on the front wheel automatically takes care of much of the imbalance. Also running with one wheel tucked under the body creates a small offset as well.... 4) One theory is that Imbalance in the rears is good in that it provides beneficial steer into the rail (I can not prove it). 5) Another unproved theory is cars that are balanced in the rears become less stable in the transition (if the front end lifts it creates an imbalance on the dominant rear wheel that steers the car away from the rail) (I can not prove it.)
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weight
May 20, 2009 4:42:34 GMT -7
Post by Cecil B. Deluxe on May 20, 2009 4:42:34 GMT -7
Is this the same thing as positioning the weight so that the vector force on both the rear wheels is the same? (presumably moving the weight left or right toward the dominant wheel?) Seems like the friction would be higher on the non-dominant wheel if you DIDN'T do this... I have tried this before and the car was dog slow. Have not tried it since.
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badger
PDDR Forum Member
[Mo0:0]
Posts: 2
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weight
May 20, 2009 19:28:50 GMT -7
Post by badger on May 20, 2009 19:28:50 GMT -7
Okay, along the same lines then, what method do you use to find out weight distribution per wheel? Is there a slick way to do this without three scales? Or without a setup where you have a surface exactly level to the scale's weight pan?
To W Racing:
Can you help me out with the math? I've never taken statics, so the concepts are a bit foreign to me. I drew out the wheel triangle (it's a right triangle), but can't seem to think how to balance the weight between the two rear wheels?
Also, I'm not sure I understood your point #2. Stacking higher? How does that reduce the car's potential energy? (I mean, I understand the benefit of moving weight to the car's rear.)
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weight
May 21, 2009 8:27:53 GMT -7
Post by W Racing on May 21, 2009 8:27:53 GMT -7
Badger,
Think about an 8 foot 4x4. Place a small 2x4 scrap under each end. If you place a 5 lb weight in the middle of the 4x4 how much additional weight does each 2x4 carry?
Now place the five pound weight directly over one of the 2x4's. How much weight does each 2x4 carry?
Now place the weight 2 feet from one of the 2x4 and think about how much additonal load each 2x4 bears
I will send more tonight.
Additional Weight on the left 2x4 = 5lbs * (distance of weight from the right 2x4 / length of board)
Plug in the numbers for the three casses above...
For #2
On a pinewood derby car potential energy is determined by how far back the weight is (More potential energy) Also having the weight lower adds more potential energy.
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weight
May 21, 2009 18:34:55 GMT -7
Post by Shade Racing on May 21, 2009 18:34:55 GMT -7
my shake a leg the lead is triangulated in it & it is the slowest car I have every race at PDDR
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weight
May 21, 2009 23:31:20 GMT -7
Post by W Racing on May 21, 2009 23:31:20 GMT -7
The formula I use is as follows. It is very simple but I believe it to be accurate.
Assume that Left wheel is raised Assume that the DFW (Touching front) = 1.0 ounces
Then Left RW = 2.5 ounces Then Right RW = 2.5-1 = 1.5 ounces
Total weight on all wheels = 2.5 + 1.5 + 1 = 5 ounces
I ran several scenarios and this works for the ones I tried.
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weight
Oct 12, 2009 9:52:20 GMT -7
Post by Bones on Oct 12, 2009 9:52:20 GMT -7
Hey Fatboy, It's a little funny that YOU would ask a question about WEIGHT. lol
How 'bout them Cowboys
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