In theory if you have a 100 inch w/b and you put jackstands at 70 inches behind ft wheels or 30 inches in front of rear wheels and the car balances there you could say that it is about a 30/70 ft to rr balance

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Turbocharged Ecotec,vanogon midshift,Berrien WoodsBuggy wagon.

Seems to me if the car balances from a center piont, ie 70 inches behind ft wheels or 30 inches in front of rear wheels, then that would be a 50/50. For a 70/30, you need to know the weight of your car. If it's 1000 lbs, that would be 700 lbs in the back and 300 on the front. Right?

if it was a 50/50 then the balance point on a 100" wheel base would be at 50" center but you are rite on the weight, at 1000 pounds it would be 700 rear and 300 front

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lets just ride

Maybe your right. My thinking was it doesnt matter where the center is, as long as the buggy is balancing, it is 50/50.

think of it like a sesaw, in order for 300 ponds and 700 pounds to balance the pivet point has to move from center

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lets just ride

Your right, duh................had my mind wrapped around it one way, but now see what your talking about.

yea, but part of the trick is what the bias is with the driver in it and the buggy under power, with the rear end squating (assuming your talking about climbing). For racing I try to get it closer to 50\50 for cornering and sliding.

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Maybe you should chug on over to mamby pamby land....

you are exactly rite racebug figuring for the driver is easy but figuring for the buggy under power i dont have the math skills for lol

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lets just ride

just guessing i think you would need to calculate tork at the rear wheels and suspension load under max torque, but exactly how to figure that i have no idea

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lets just ride

EXACTLY. Thats why giving people generic dimensions (like some were asking for in other thread) will do them no good. I guess it would be a starting point, but thats about all it is if you are trying to build a good light climber. The lighter it is , the more small weight and balance changes will affect it. There is no rigid formula.

ok does anybody know aprox weight bias of your car sitting still on the ground with driver or driver/passenger in it. My car is front end heavy compared to most i've seen. Using the generic jack stand method my car with winch and battery up front and two 200lb people sitting in it i have to place jack stands almost centered between my wheels to get car to balance. This is a stalker chassis with an ecotec engine and 103' wheelbase with 3x3 arms. I like having some nose weight but not that much. I moving my battery to the rear behind the seat and may move some more weight around and i'm curious as to how that compares with other buggys.
Thanks Tony

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Turbocharged Ecotec,vanogon midshift,Berrien WoodsBuggy wagon.

Ours is 42:58 front to rear with driver and passenger in and a full tank of fuel (tank is in the front, as is the battery and power steering pump)

The folks that can do all that math make enough money to buy the best chassis money can design and they don't have to do all that math. Isn't it ironic?

Find a circle track guy in your area that looks like he has one a race or two and he will probably have a nice set of race scales. Bribe him with beer to borrow them for an hour or so and you will know the static weight bias. Then you can really start doing some tuning. If you have a fairly adjustable suspension system you will be able to play alot with how the weight gets shifted around under power. You don't have to be a rocket scientist to get close. You just need to be willing to invest the time in tinkering.

For balancing a buggy there is a simple relation that will help. It defines that the moment(torque) is proportion to length time’s weight.
The calculus relation assuming mass is constant is: M(l)=∫Wdl where w=weight and l=length. With a rigid arm like a buggy chassis that suggests l is constant and the intralgral relation will form an algebraic equation as described above. Where M=WieghtxL
Just as the previous example mentioned if a 300lb person and 700lb person were on a teeter totter and the length of the totter was 10’ then the respective lengths would be 7’ (300lb) and 3’ (700lb) Therefore, both moments/torque would equal 2100 ftlbs

700lb 300lb
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<-------3ft------><-----------------7ft----------------->

The net torque would equal zero

I think it would be safe to say out of the 1905 members here on Woodsbuggy, you just lost 1900 of them. Myself included. The only thing that saves my bacon is too many years foolin with dirt cars setting up chassis.
Keep this in mind, everything we do is a series of compromises. You will not get things perfect no matter how hard you try. Find what works best for you and make the best of it. In dirt track racing a good driver can adjust his driving to make up for a bad chassis set up, same thing here.

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Of all the things I've lost, I miss my mind the most.
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Sometimes I have a hard time explaining things, but you are probably right. I apologize for the confusion. It as simple as this: If you hold a weight with your arms extended the heavier the weight the harder it is to hold, and if you bring your arms in the easier it is to hold. So the only conclusion to be made is that you can add less weight if you move the weight further from the center of mass.

We used to race late models and modified's. Its been a while but I would agree with what you said.

I will try and keep the math out of it next time... LOL.

Thanks

So do you know the weight bias of your car SITTING STILL and LOADED?????
Thanks Tony

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Turbocharged Ecotec,vanogon midshift,Berrien WoodsBuggy wagon.