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wobblywalrus
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« Reply #285 on: November 02, 2010, 01:04:13 AM »

Thanks for pointing this out, Stan.  And the help is appreciated, Mr Observer.

The dyno curves for torque and horsepower are plotted along with vertical lines that represent the down and back runs.  The engine speeds are estimates from the RPM vs Speed Chart.  This is a reconstruction of this year's runs and it gives me the info I need to go faster next year.

Many posts ago, at the beginning of the engine tuning, I mentioned 7,500 as my target rpm.  This is the engine speed I want to have at full throttle through the measured miles.  It is a compromise that gives me reasonable performance and affordable engine life.  I was running right at my target rpm.  Any more speed will cause the rpm to increase above the target and it will shorten engine life or worse.  Task 1 will be to have 40 and 41 tooth rear sprockets cut.  This will allow me to go faster at reasonable rpm.

The run curves also show that my torque is dropping off rapidly when I reach maximum speed.  Also, I am past my horsepower peak.  I need to move my torque peak closer to 7,500 rpm.  Not much, about 500 to 700 rpm closer would be great.  I need to be careful here.  I need a broad spread of torque, with midrange torque to get me to top speed and high end torque to make good power when I get there.  Some intake and exhaust tuning will do the trick.  These are Tasks 2 and 3.

These run curves also show me the limits I have.  Maybe, with hard work, some help, and luck, I can get the horsepower into the mid to high 70's.  That is not much.  Very little, really.  I need to work hard on aero.  That is Task 4.  Lots to do.  Fortunately none of this is really expensive. 

   


* 2010 Run Curves.jpg (227.37 KB, 600x780 - viewed 174 times.)
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« Reply #286 on: November 03, 2010, 07:43:42 AM »

In support of near-namesake Interested Bystander, he, as an admitted “reader”, could have done just as well since it was just a matter of reading through the documentation.  Wobbly was looking for bark beetles and had forgotten which forest he was in.
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wobblywalrus
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« Reply #287 on: November 04, 2010, 01:09:34 AM »

This year was the biggest race of my life and it was time to be prepared.  So I brought all sorts of wrong size sprockets, left most of my tools and spares on the workbench at home, and I did not have dyno curves for this year's engine.  It was a triple crown of dumbness.  Usually I am prepared and this is what I do.

First, I estimate the engine and wheel sprockets that will work best.  Then I make a run through the measured mile.  Now I have a time slip with a speed.

Second, I do my best to estimate the engine rpm.  A record from a data logger is best.  Visual tachometer readings are second best.  Calculating the rpm based on estimates of slip and the "RPM vs Speed Chart" is a last resort.  I always have the chart, just in case that I will need it.  Now I plot up that first run's rpm on the dyno curves for the engine, as shown on the "2010 Run Curves."  This tells me the engine's torque output during the run.

Third, I calculate the driving force at the rear wheel contact patch using the speed on the time slip and and the corresponding engine torque.  See attached "Driving Force" paper.

Fourth, I calculate the driving force at the speed on the time slip with the gearing change.  This could be a sprocket or tire change.  The proposed change is justified if it significantly increases the driving force.  An example will be posted tomorrow.

Most of my formulae have multipliers and these are for my bike, only.  They make it easier for me to calculations on the salt with a pencil and paper.  The RPM formula on the "RPM vs Speed Chart" has a multiplier of 1976.  The "Long RPM vs Speed" attachment to an April 2010 post shows the full formula without the multiplier.  The "Driving Force" formula has a 44.9 multiplier and the full formula is shown on the sheet, too.  A person wanting to use these methods needs to calculate their own multipliers.  Also, I do not correct the torque to Bonneville weather and climate conditions when I am making simple comparisons.  The correction factor, when I use it, is 86 percent.  In other words, torque and horsepower at Bonneville is 86 percent of what we measure on the dyno in Beaverton.   



 


* Driving Force.jpg (227.5 KB, 600x791 - viewed 177 times.)
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« Reply #288 on: November 04, 2010, 01:46:02 PM »

WW,
86% seems to be a good number.
I have been using it for jetting [I'm at sea level] ever since Leroy Newmeyer told me to adjust 14%.
For me if it was not right on it was darn close.

Don
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« Reply #289 on: November 04, 2010, 09:43:22 PM »

Wobbly, a couple of musings on your power curve---

In “observing” your power curve in reply 287 the “flat-topped” horsepower curve just didn’t seem consistent with the torque curve.  After cranking a few numbers, it appears that the horsepower data points at 7400 and 7800 rpm are plotted one unit too high.  While this is not a big deal, correcting the locations more vividly shows that you were running well beyond the approximately 6800 rpm current power peak.

Also, as you embark on more inlet and exhaust tuning, it may be useful to recall that what may be good at 60 degrees in Oregon may not be quite optimum at 95 degree Bonneville.  This would amount to about a 3.5% increase in acoustic celerity--again, not a big deal, but an indication of which way to round off dimensions, or a reason to include some “shimming” capability in the inlet tract. 

Getting the inlet, exhaust, and valve timing all coordinated will probably keep you well occupied over the winter.



Re: Your “driving force” performance evaluation process (reply 289):

First, there is a much easier way to calculate the tractive effort, or “driving force.”  Force times velocity equals power.  More particularly,  Force (lb) x Velocity (mph) / 375 = Horsepower (hp)  and conversely, Force (lb) = 375*Hp/mph.  (This is just a linear version of a relationship you are already familiar with--Hp = torque (rotational force) x rpm (rotational velocity)).

Example:  70 horsepower expended at 125 mph produces 375*70/125 = 210 lbs of “driving force”.
One could also include an efficiency factor to account for driveline losses.

Second, why do you care what the torque or driving force is,  except to get an idea of the aerodynamic drag?
All the torque in the world won’t pull the skin off a grape.  To do work (go 130 mph at Bonneville) one must apply POWER.  Likewise, to accelerate you need to apply power.  Don’t worry about how broad the torque curve is or is not, worry about the power curve.  Manipulate the power curve itself by tuning changes, and its application by gearing arrangements.

Of course, none of this is meant to criticize you or your methods, it just illustrates a different and perhaps simpler perspective that may be useful.
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« Reply #290 on: November 04, 2010, 11:46:00 PM »

Thanks for finding the error in the horsepower curve.  Some of this is late night work and mistakes are easy.  The torque method seems to be easier for people to understand when discussing the effects of sprocket and tire size changes.  It gives similar results to the power method.  I used the power method to verify the results of this torque method before I posted it.  I can post the power method results, too, if requested.

Let's pretend I came to the flats with extra 41 and 40 tooth rear sprockets.  As seen by looking at the power and torque curves, the overall gear ratio is high and I am running well beyond peak torque and a little bit past peak power.  Do I put on a smaller rear sprocket and make another run?  Will it help?  This is a big concern.  The engine is torn down every ten runs for inspection and this is expensive.  I do not want to waste a run while using an idea that might not work.  It is time for some quick figgering.

First, I use calculate the driving force that gets me to the speed I am running.  In this case it is 192 pounds force.

Second, the rpm is calculated that will occur at the speed I am currently running with the proposed gearing combination.  The engine torque is figured out for this rpm using the dyno curves.

Third, the driving force is calculated for the proposed gearing.  It will make me go faster if it creates more driving force.

I initially ran a 42 tooth rear.  Calcs are done for a 41 tooth rear and it would give 194 pounds force.  A 40 tooth rear will give 195 pounds force.  These are small increases and not worth the time, effort, and wear and tear on the motor. 

 


* Sprocket Change Calcs.jpg (231.31 KB, 600x794 - viewed 173 times.)
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wobblywalrus
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« Reply #291 on: November 06, 2010, 11:57:30 PM »

Sprocket change calcs using the power method are attached.  This procedure is explained in the book by John Bradley "The Racing Motorcycle: a technical guide to constructors"  ISBN 0 9512929 2 7

The torque and power methods give me slightly different results and the power method's are usually higher.  The torque method I can verify to be correct by vector diagrams and the principles of simple machines.  I use the torque method because of this.

Either way, the methods show that a sprocket change would not be worth it.  I was hoping it was what I needed.  It appears that I need more horsepower. 


* More Sprocket Calcs.jpg (230.5 KB, 600x794 - viewed 172 times.)
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WhizzbangK.C.
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« Reply #292 on: November 07, 2010, 12:23:39 AM »

Sprocket change calcs using the power method are attached.  This procedure is explained in the book by John Bradley "The Racing Motorcycle: a technical guide to constructors"  ISBN 0 9512929 2 7

The torque and power methods give me slightly different results and the power method's are usually higher.  The torque method I can verify to be correct by vector diagrams and the principles of simple machines.  I use the torque method because of this.

Either way, the methods show that a sprocket change would not be worth it.  I was hoping it was what I needed.  It appears that I need more horsepower. 

Using the HP method, you show to have 11 more pounds of force with the smaller sprocket. In my mind that's enough difference to go for the sprocket change and give it a try just to see how much more speed that might actually give you. Have you calculated even smaller sprockets, say down to 35 teeth or even smaller? You may find that there's a sweet spot somewhere in there that would justify hunting one down or even having one made if not available, even if you had to run it out in a lower gear. Seems like a lot of folks don't pull top gear and are setting records, reference Ack Attack.

You've mentioned in the past working on your aero package more, and I think your calculations point out the importance of that also. Better aero will decrease the driving force required to achieve a given speed. Since you've established how much driving force you have available with your current engine package, perhaps you can now calculate how much drag you need to lose to get to your target speed, and work your aero mods based on that?

On our 250 bikes, we seem to get the best results running smaller sprockets in 3rd gear (4 speed gearboxes). We've experimented with larger sprockets and trying to pull 4th, but haven't had much success due to the large jump between 3rd and 4th, just a characteristic of the design that we have to deal with and may not apply to yours at all, but I thought I'd mention it as something you may want to crunch some numbers on.
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« Reply #293 on: November 07, 2010, 04:57:06 PM »

The disparity in tractive effort calculated between the power method and torque method is due to the necessity of using the tire circumference in the torque method.  All the rest of the analysis is completely consistent.  It is not clear how Wobbly determined the 79.25” dimension for this circumference--whether it is the unloaded circumference or a loaded rollout measurement.  Since reducing the loaded rolling radius from that of the 79.25 circumference by .63” (a likely amount of tire deflection) brings the tractive efforts into agreement, it would seem that the 79.25 figure is unloaded, or otherwise slightly erroneous.

Also, rather than making various stabs at different gearing, if, in this case 70 hp produces 129.5 mph and most of the drag is aero (which goes as power cubed) and the maximum power available is 72 hp, then P2/P1 = (V2/V1)^3 and V2 would equal V1 times the cube root of the power ratio, or 129.5*1.0094 = 130.7 mph.  If the bike were re-geared to operate at the maximum power, it should be able to produce the 130.7 mph.  Then the question is whether a 1.2 mph increase is worth it.  This seems more informative than just determining the change in TE.  (On the same basis, 80 hp would produce 135.3 mph--is that worth it?)
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« Reply #294 on: November 08, 2010, 10:31:14 PM »

Thanks, Mr Observer.  I was using the unladen tire circumference measured with a tape.  That explains the difference.  You are smarter than me.  I spent an evening trying to figure out why both methods gave different results and I could not figure out an answer.

K.C., valve size is a problem in these little Triumphs.  The intakes are 2mm larger than standard.  This change, with a port job, really helped the 790 cc engine.  This bigger engine can benefit from 4 mm larger intakes and exhaust valves.  South Bay Triumph makes these.  I am not sure if I will put them in.  Lately I have been riding around town on frosty mornings, in the rain, at night, and there are leaves all over the road from the street trees.  This engine is very well mannered and I can tractor around in these conditions without wheel spin.  This is the best street engine I have built, by far.  It is hard to make it better and easy to screw it up.  Like you say, it is time for me to work on aero.

Talk among land speed racers can be sorta casual about blowing apart motors.  Experience has shown me that an engine failure leads to a series of events that I do not control.  One of these is the classic "high side" when a person is flipped off the bike and into the air.  Engine reliability is especially critical on a bike like the Triumph.  The spinning transmission shafts and gears are inches away from the rods and pistons.  Parts from a fracture can easily fall into the tranny and lock it up along with the rear wheel.  Life would suddenly be very interesting.  This is what I do to lower the chances of a blowup.

First, the bike has a rev limiter and it is set to an engine speed about 500 to 1000 rpm less than "explosion."  It is hard to look at a tach and use the throttle to limit engine rpm on the salt.  Too much is happening too quickly.  A rev limiter is essential.

Second, the engine is built to be strong using the best parts available.  The motto is "built it stout now and worry about performance later."

Last, there are periodic tear downs and inspections.  These attempt to find potential problems before they become disasters.

The reason I mention this.  The tire broke loose twice during the down run this year.  The engine hit the rev limiter both times before I rolled back on the throttle.  Last winter I did a periodic tear down, problems were identified and fixed, the bike has a rev limiter, and it is set to a safe rpm.  There is a good possibility, had I not done all of this, that engine would have blown.  As they say, be careful, be safe, and go fast.         

 
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« Reply #295 on: November 09, 2010, 12:51:59 PM »

Off the basic subject, but it's a question that has always evaded an answer from me:

"I was using the unladen tire circumference measured with a tape.  That explains the difference.  You are smarter than me.  I spent an evening trying to figure out why both methods gave different results and I could not figure out an answer."

I can understand pretty well how the diameter of a tire will change between loaded/not loaded -- but I don't see how the circumference can change.  Whether squished down harder on the surface or not, there's no change in the amount of rubber -- it doesn't fold in or anything, so 100% of it still has to travel on the surface.  It flexes from round to flat and back, sure -- but it doesn't grow or get shorter.

Please teach me how the circumference changes -- or tell me that I really do have it correct.  Thanks.
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« Reply #296 on: November 09, 2010, 02:10:32 PM »

Tire under load squishes down resulting less height from top of tire to ground - smaller dia.

Smaller dia divided by 2 for new radius multiplied by pi [a constant 3.14xxx] results in a new smaller "real or effective" circumference.

Less circumference means less "roll out" or forward distance gained per revolution.

Help any?

Nearly 40 years ago as I was changing eng sprockets [countershaft sprocket wasn't handy to change behind the clutch basket and rear was one piece with the brake drum] looking for optimum gearing a fellow watching made the comment that the final ratio may be the same but it makes a difference where the changes are made, ie between the eng sprocket and clutch basket or between the countershaft sprocket and real wheel sprocket.  In other words, where the torque is multiplied in the chain of events can make a difference even though the numbers "should" give equal results.  Just now starting to understand that concept myself.

                      Ed
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« Reply #297 on: November 09, 2010, 03:17:35 PM »

SSS,
The essential dimension for use in Wobbly’s torque formula is the effective rolling radius--the distance from the center of the axle to the ground plane.  This is approximated in his formula by calculating from the tire circumference, presumably because that is (arguably) easier to measure. 

So, you are mostly right that the circumference itself probably doesn’t change substantially.  However, being of rubber composite construction, the circumference probably does change to a small degree when it is deflected by the ground, although the shorter (straight) distance along the contact patch is mostly made up for by slight outward bulging of the tire ahead of and behind the patch.


Ridgerunner,
Don’t give much creedance to your onlooker’s comment.  The only difference would be in friction losses, possibly due to the type or number of components in the drivetrain.
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« Reply #298 on: November 09, 2010, 03:54:35 PM »

IO, I'lkl take your response as about what I had figgered on my own - and thank you for corroborating my own story.  I can see how the effective radius - axle to ground - is made smaller by the tire's deflection, and therefore how that'd affect the torque.  And this way I'm allowed to keep my circumference about the same (I still think it doesn't change at all) because the amount of rubber doesn't change -- save maybe for some small amount due to the outward "squish" of the tire.  Thanks to both of you.
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Jon E. Wennerberg
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« Reply #299 on: November 09, 2010, 10:56:55 PM »

I would have thought that at, say 120 mph, the centrifugal force exerted on the tyre would have caused it to grow even when it had some load on it. I know that with some of the custom stuff i used to work on, and in particular older scooters, if you put a slightly larger rear tyre on, it would spin on the stand with the wheel in the air but as soon as you got some speed up on the road it would run into the engine case.
Having said that, it would generally be "J" and "P" rated tyres that were not really high performance equipment
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