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OK you math wiz's

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BuzzStPoint

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Working on a Monster truck project and need some help.

Chain and sproket movement..

If the object were to move 30feet, I need a box to move 35 inches.
Understand?

for every 30 I need the sprockets to pull a chain for 35 inches.

Anyone know the formula to figure that out?
 
Sounds like he is trying to make a Monster/tractor Truck Pull. Something like you would see on TV with Monster/Tractor trucks.

Which would mean he is looking to run his truck 30ft with a contraption on the back that has a box pulled by a chain and he wants the chain to pull the box towards his vehicle 35 inches.

Hope this helps. And if I am completely off kilter feel free to tell me...=)

Later
Whistlre

(Saw one on E-bay the other day for like $200.)
 
Originally posted by Whistlre
Sounds like he is trying to make a Monster/tractor Truck Pull. Something like you would see on TV with Monster/Tractor trucks.

Which would mean he is looking to run his truck 30ft with a contraption on the back that has a box pulled by a chain and he wants the chain to pull the box towards his vehicle 35 inches.

That is correct.

why is this in the t-maxx forum?

Cause I have this forum in my favorites, so when I click, it just comes up to this forum.
 
This is going to be dependent on a few things. One would be the sprocket size you intend to use (ie diameter). The type of chain you in tend to use (ie bicycle chain or other style). The rest is going to be dependent on the drive mechanism as in what is going to make the chain actually move. Since I am not entirely familiar with how a sled works in real life, I would have to look at one design to give you a best recommendation.

The basic math will translate to the number of teeth on the sprocket times the length of a link that will ride that sprocket. This will give you the chains travel through one rotation of the sprocket. From there, you will need to figure out how many revolutions add up to 35inches. That is as simple as dividing 35inches by the number of inches the chain travels under one rotation of the sprocket.

Beyond that kind of math, you will need to figure out the workings of your mechanism and that requires a sled design.
 
Yup, it's all about the ratio. But to throw my .02 cents in, here goes.

If you convert the track length into inches (30' = 360") and the sled length is 35", then the ratio would be 35/360 or 10.29:1.
So, for every 10.29" the sled travels down the lane, the weighted box would move 1". So, if my math is at least close, you need to set up your gearing in the 10:1 range. If my math/figuring is completely out to lunch just slap me.
 
We may just slap you for the sake of it, but your math seems correct.
 
A very simple formula that was in 9th grade algebra.

I wish I could help you, but it's the only class I ever failed.
I'd trust Militarymaxx on this one. It looks totally confusing to me, so it must be right.
 
in 9th grade I still wasn't in algebra...lol So dont ask me, but a tractor pull with a Maxx sounds kinda fun, but Hot!
 
Originally posted by militarymaxx
Yup, it's all about the ratio. But to throw my .02 cents in, here goes.

If you convert the track length into inches (30' = 360") and the sled length is 35", then the ratio would be 35/360 or 10.29:1.
So, for every 10.29" the sled travels down the lane, the weighted box would move 1". So, if my math is at least close, you need to set up your gearing in the 10:1 range. If my math/figuring is completely out to lunch just slap me.
Dude, you get major kudos for that one!
 
Another factor will be tire size on the sled. The rollout will determine the average number of revs the tires will turn in your 30' run. Then you'll need to base your ratio on how you're going to move the chain 35" in that many turns of the wheel shafts..
 
OK here's what I do know.

The sled is 42 inches long.

The weight box will travel 30 inches

Tire size is 7 inch.

Thats all I know.

So I now have to move the weight box 30 inches in 30 feet.
 
If you make your tire size the same as your drive sprocket there will be no problems. It won't change your ratio.

VB
 
But I dont have any sprockets yet.

This problem is what I'm trying to figure out. What size sprockets I need.
 
Umm....I might be wrong, but if you need to move a box 30" in 30', wouldn't that be a perfect 12:1 ratio?
 
So if he uses a gear on the axel that is 7" in diameter, that cancels out the wheel diameter factor. Then he just needs a big gear that has 12 times the # of teeth to gear it down. But, I wonder how big that gear would be. Might have to do a multi-stage gear reduction...
 
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