calculation of top speed?

T

TJ500

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Joined
May 17, 2003
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661
Location
Lockport, Manitoba. Canada
anyone know off hand the calculation for top speed (mph)... Need the formula that takes into account the track thickness too...

also need diameter or circumference of 9 tooth drivers used in calculation..

(couldn't find in Tech pages)

thanks..
 

Okay I'm up to the challenge to wing it and then look it up later in aaen's cltuch book.
V (velocity in mph) = w (revs/min ala rpm) * r (radius of the tire in effect) * C

w=engine rpm * clutch ratio * gear ratio
r= radius of driver (where it meets track in inches) + track thickness (where it meets the driver to where it meets the ground (top of lug on hard surface) in inches)
C=conversion of in/min to miles/hr & angular velocity to linear (rad to rev) = .00594
This all assumes no silp. Clutches can shift 1-1.1 I believe.

I'll post aaen formula later but they should be the same. Jason
 
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Tj, You Take Rpm{ We Will Use My Sled For Example} 9200 X My Gear Ratio For 660ft On Ice, 23/39=0.58 X 1.89 X 0.01136 = 114.56 Mph At A 1.1 Ratio.
Now On The Asphalt That I Have Found If Your Helix Is Cut = 10 More Mph At Full Shift Out. With 9t drivers.
 
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I would think to be "dead nuts on" you would want to just 1/2 the thickness of the belt of the track.
 
I use......example my 02 Viper

rpm (8500) X 3.5 (radius of drive sprocket) divided by 1.73 (gear ratio of a Viper) divided by 168= 102.36 (theoretical speed with clutches at 1 to 1 and no slip). Then multiply by 1.05 (commonly 5% overdrive) 107.5 MPH. This is usually within 2 MPH of the Stalker on a hard pack lake condition. My Viper ran 109.2. Clutches usually will give you more than 5% over but with track/belt slippage and such I figure in the 5% and I get a more realistic figure. Trying to figure track thickness and such as it may give you an actual theoretic figure means nothing on snow.
 
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This topic always creates some emotional responses. I'll still bet on my formula rather than multipling by this and dividing by that. You guys are all right that a snowmobile always has slip with the snow and you cannot tell exactly where the contact patch meets the ground. But the tip speed of the track is going to be v in my formula. Whether or not the snowmoblie is traveling at v miles per hour is another question due to slippage.
Anyways plug in the agreeded numbers into my formula: rpm=8200, clutch ratio=1.05, gear ratio (21/39) and the driver diameter of 3.5 and you get 99.9mph. There are other ways to get the effective radius of the tire (track) by using its pitch.

However you must add to the radius of the driver the distance where you think there is no/least amount of slip to do the calculation. I would bet the belt thickness is close. So lets say the drive is 3.5" and the belt is 1/2" thick then the speed on the outer edge of the belt is 114 mph. If you wanted to calulate the speed at the lug tip of a one inch track it would be 3.5" +1" =4.5" or 128 mph. Is 3.5" really the radius of the drivers?

I did have a small error in my conversion factor above that I changed. But I will now bet on my formula rather than multiplying this by that and dividing by another. Man I hope my two engineering degrees are not going to let me down!
 
I do list in my formula the point at which I think there is the least chance of slip. The point at which the driver interlocks with the track. Anything after that is a gray area. Trying to figure out track slip and belt slip and other issues (coefficient of drag)is a waste of time, nor anything that could be measured consistently.
I'm sure your 2 degrees won't let you down. Unfortunately, I don't have any (but I do have 25yrs. of snowmobile tuning under my belt). What I do have is a Stalker STATS system and a laptop which I do my tuning with. And yes it is more acurate than ANY of the formulas which you can sit a home and wish how fast your sled will go. All I stated in my post was how the generic formula I presented relates to what I see in the real world.
 
800 I'll agree with you 100% percent that no one could accurately and repeatably predict the speed of a snowmobile from their living room. The theory goes out the window when your butts on the seat! I was not knocking your expertise either..

I think your formula and mine might even be the same other than your are using the inverse of the gear ratio and dividing, which is the same and I'm not sure where 168 comes from. In fact if I use your numbers 8500*3.5*(23/40)*1.05*.00594 I get 106.9. My point is you have to know how the 168 is derived in order to make sure its truely a constant and applicable to other situations.

It would be interesting to see what happens on the STATS system when you drop or add a tooth..

On this same subject BMW/Williams was able to predict the pole speed within a tenth of a second at the Chinese grand prix before there were any cars on the track. So I gues just about any is possible with enough money and data. But I don't any of us on here or even ty memebers combined has enough.
 
thanks for the help guys... Absolute perfect accuracy is great but not really necessary for me... Weather I'm travelling 112 or 114 mph down my driveway and out of my yard on my sled really isn't important, what's important is simply knowing the wife won't be able to swing the broom quickly enough to hit me! HA HA..


thanks again TJ..
 
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The 168????? I got that from an old snowmo tuner years ago, don't know where he got it or if he made it up, all I know is, it works. Some things are just that way.
 
I've got lost in my own tread!

Anyways 800 it looks like you have a version of Aaen's formula which for all intensive purpose is the same one that I winged out there the first time. Here's aaen's:
MPH=(engine speed * sprocket pitch diam)/(shift ratio * gearing * 336)
engine speed is in rpm, gearing is driven sprocket/driving sprocket, sprocket pitch=sprocket diameter + track thickness in inches.

Your 168 is half of 336 because you enter the radius instead of the diameter. I think we both have the same frickin formula except we are getting caught in significant digits of the conversion factor.

An intersting thing in his book is he says the clutches are most efficient at 1:1. So why are the other companys selling overdrive sheaves? Because of this efficieny he says to gear tall to take advantage of the efficiency which drops off in overdrive.
 
the book

I always wanted to get one of those books. it's true clutches are less effcient as you go past 1:1. As the overdrive increases so does the load required to push the belt up. Actually, it doubles with the overdrive numbers. So thats why I always try to set-up for 1:1 numbers and anything you get past that in smoother conditions is gratis. That again is why I figure the 5% number and if you see a big number on a given day, just consider it a good day. Unless you've got major power, relying on overdrive and overdrive kits is a waste of time. Thats why alot of times when you get an overdrive kit you're told to gear down. Why, because you don't have the power to use what you put on for what you think you did. Overdrive kits give you a wider effective ratio, meaning, you can gear down and get better accelleration while keeping the same top end. Lots of people buy those things thinking they'll install them with the same gears and pick up top end. Lots of times you'll go slower.
 
I have that book also,and read the same thing about clutch efficiancy in overdrive,that is why I asked a question the other day on here about actually gearing up 1 tooth on the top gear to get the clutch to run in a more efficiant mode.-Harvey
 
Harvey most of our sleds are over geared as they are. Use the formula and figure out the speed of your sled. Is it capable of going that fast? Probably not if its a stock sled. The problem with gearing up and not having the power to use the gearing is, generally you will go slower. Remember, when you gear up, you're changing the load on the motor from the start, and if you don't have enough power to use it, it won't go as fast as it did before.
 
Lots of talk about this, If you want to go fast on the lake, gearing is the way to go! It really won't kill your bottom end, below 25mph you might be able to feel it, above that the clutches actually shift out less, keeping your gear reduction total the same. Overdrive efficency is terrible, don't go there, some days if conditions are perfect, you will get a little bonus. Sometimes you'll pick up some acceleration cause you'll be loading the motor down more, kinda like going to a steeper angle helix. lots of ice racers in the know, will actually gear for higher top speeds than they actually need. If your spinning your track alot at low speeds, then more ratio is useless. A stock viper is geared for 106 MPH, stock 1to1 ratio, not that out of the question for horsepower to weight ratio. Clutching may have to be fine tunned, price for having the best of both worlds. The exception is of course mountain riding, where horsepowers are reduced, and speed is not of concern. I have geared sleds both ways, and this is my observation from experience. And yes evey sled I have owned was capable of easily reaching 1 to 1 clutch shiftout. You'll just have to try it and let us know what you find out.
;):D
that little bastart looks like the old lady last night, Oh yea she wasn't even drinking, you figure it out!
 
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