Landracing Forum
Tech Information => Technical Discussion => Topic started by: isiahstites on October 20, 2007, 12:09:00 PM
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I know the with the speed given by the dyno the bike is not moving any air and probably not the same load as you would see at the track, but I am curious how far off is the dyno speed versus what you have seen at the track?
Scott
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figger 20% + or -
kr
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Not sure of the question. Are you asking if the air speed on the salt affects power in a way the dyno cannot simulate?
Yes.
Or are you asking if the RPM change used to record power on the dyno is accurate and linear compared to the actual conditions at high speed?
No. The typical dyno is programmed to permit engine speed to increase by either 300 or 600 RPM per second (other settings are possible), since that most closely mirrors the transients experienced in racing venues where acceleration is the key factor. Larry Widmer has frequently commented (paraphrased slightly) that peak flow CFM and peak power are not the crucial goals, and that an engine with high transient power will outperform the former in most venues.
IMO acceleration in low gear (rapid RPM change) on the salt is of minimal comparative value vs. achieving top speed, where engine speed change will be very slow and transient power less important.
Obviously, every properly geared vehicle is straining to pull high gear (viz. acceleration almost stopped just before the traps), but only the faster vehicles are really stalled in high gear because of their extremely high wind resistance
This is not the dominant factor in slower vehicles where force × mass has a high value, and is only present for a few yards in drag racing and never in sprint cars.
I don't know if the faster cars are using dyno pulls more like 100 RPM/second, but it might generate some interesting data. I would be looking at plenum volume, the trade-off of CFM for port velocity, header primary diameter, and flow-through on overlap. All these are compromises with values favoring best acceleration will frequently conflict with values favoring peak power. Sizes and shapes that move the torque peak up will have greater positive effect on top speed than the loss of acceleration, especially where traction is limited for most of the run.
The very high load and absence of residual accelerator pump discharge also means that some engines will benefit from smaller air correction (curve it rich rather than jet it rich) to reduce chamber temps, and some spark reduction for the last mile (perhaps 2-4°).
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Just wanted to know an estimation of how far off dyno speeds were versus track speeds since you have wind resistance on the track.
Scott
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"Yup , just like they said and said and said, a dyno might float yer boat, but it won't teach you how to sail." (me)
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Scott is using a Dynojet inertia dyno, which isn't programmed for any particular acceleration rate. It just has a big drum of a known mass moment of inertia riding on pillow block bearings, and a simple optical sensor that tells the electronics every time it completes a revolution. The electronics calculates the horsepower from the speed and acceleration rate of the drum, knowing it's mass moment of inertia. Since the circumference of the drum is also known, it can also calculate ground speed. If a tach pickup is used, it can also calculate torque, since it knows horsepower and engine rpm. It's kind of a weird torque number though, it's engine torque, i.e. upstream from the gearing, but including drivetrain losses.
So a first gear dyno pull will happen really quickly. A top gear dyno pull will take several seconds. Results are similar in each case, because even though the acceleration rate varies, so does the speed range. High acceleration at low drum speed (i.e. a first gear pull) indicates the same horsepower as low acceleration at high drum speed (i.e. a top gear pull).
Anyway, Scott, on my Dynojet, the speed is a little off from what I would get turning that exact same rpm in that same gear at the Salt flats, primarily due to wheelspin. Wheelspin is essentially zero on the dyno, it's never zero at the salt flats, at least not at the speeds and power levels of my bikes. Maybe for a sub 150mph bike or something you can ignore wheelspin.
The other thing about wheelspin is that it's different every time I go out there. I use a spreadsheet to calculate gearing and choose sprockets. But until I make the initial pass and get a handle on how much wheelspin I'm getting, I don't really have my arms around it.
I would advise that you gear your bike for the class record, at the power peak rpm, or maybe slightly higher, based on some wheelspin assumption. Start at maybe a 5% wheelspin assumption given the power level and speed capability of your bike. but depending on the weight of the bike, and the salt that particular day, it'll be different.
I can send you a copy of the spreadsheet if you'd like. It's in Excel.
Don't be afraid to gear it slightly short. The physics would indicate that you want to the motor right at it's power peak, but historically, my bikes have always wanted to be geared slightly deeper and taken slightly past the power peak rpm. That's where they seem to get their best speeds. YMMV.
You certainly don't want to be below your power peak rpm when the bike runs out of poop. That really makes for slow speeds, in my experience.
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I would like the spreadsheet you can send it to stites@cox.net
Aaron, since you are talking about gearing I have a question? I am currently geared at 2.0, 24 front 48 rear and the stock gearing is 1.9 or 1.8 if I recall correctly. I thought 2.0 would be a good place to start to see if the bike can pull the gearing rather than be to optimistic and work my way down in teeth on the rear sprocket. Your thoughts??
Scott
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"Yup , just like they said and said and said, a dyno might float yer boat, but it won't teach you how to sail." (me)
I'll tell you what, I've found the dyno to be an enormously valuable tool for getting prepared. Not only does it help you get the tune spot-on, and work out things like exhaust configuration, intake tract length, and so on, but it also has a way of shaking out bugs. I'd much rather have something go wrong on the dyno, at the shop, than at the salt flats. I can't tell you how many times the dyno has shown me something that would've been a problem at the flats.
No, it doesn't get you 100% of the way there, but 90% easily.
One of the big things it doesn't do for you well is show you fuel delivery issues. A carb bowl may run empty when you lean on it for miles, but you never see that on a 5 second dyno pull. I fought that at Finals. Finally figured it out just as the wind came up.
I've found that my bikes want little or no difference in jetting between the dyno and the flats. 'Course, I live at 5400'. Seems like every time I try to fatten it a tad to go down to the flats, I end up putting it back.
The dyno helps me get the timing spot-on, too. I establish where the power falls off on each side (adv & ret), and center it right between them.
I see so many guys attempt to tune at the event. That's just really hard to do, because to get feedback on a change, you've got to wait in line. I find it's much better to have your shit together just as well as possible when you roll it out of the trailer, and then go into a fine tuning mode as quickly a possible.
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Scott, I can't answer that question without a whole bunch of other information. Then I plug all that into the spreadsheet and let it tell me what sprockets to use.
Engine rpm is critical. I've got to know the engine rpm of the power peak, or better yet, a range of rpm over which it makes essentially the same power.
Then primary ratio. Yours is probably 1.6, with a 35 on the crank and a 56 on the clutch basket.
I'll assume you want to run in 5th, which is 1:1. Your gearbox has the fewest losses in 5th.
Then front and rear sprockets.
Then rear wheel circumference. I like to measure this by putting the bike on a swingarm stand and taping a piece of wire to the tread, dead center. Wrap the wire around the tire and cut it off at the exact circumference. Now peel it off and measure it.
Finally, wheelspin assumption.
From all that, you can see the resultant speed, or range of speeds over the rpm range. The spreadsheet lets you put in different sprocket combinations and see resultant speeds.
The spreadsheet will also calculate horsepower required for a given speed. This calculation is based on the cubed relationship between hp required and speed increase. So you have to put in a reference point, i.e. my bike can go 160mph on 130hp. It calculates hp required for other speeds from that reference point.
Let me throw it together with numbers from your bike and send it to you.
Can you tell me the power peak rpm range, and also the rear tire circumference?
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Let me throw it together with numbers from your bike and send it to you.
Can you tell me the power peak rpm range, and also the rear tire circumference?
Aaron,
If by power peak rpm range you are speaking of what rpm the motor made its highest hp numbers it was at 7300 +/- and the rear tire circumference should be very similar to Susan's bike as it is the same rim as hers and I have on the stock z rated tires which are 24.5" tall.
Thanks,
Scott
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We'll know this weekend how our dyno speed compares with track performance. We know we made 622hp/559tq on an engine dyno in the same configuration and ran 179 at Maxton last week. We'll be running Sat morning on a Mustang chassis dyno to see how close the numbers are.
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Holey toledo, just wandered on 2 this post and realized you are talking about MOTORCYCLE horsepower!
Mercy sakes alive!
Wouldn't send my kid out on that rocket!
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Holey toledo, just wandered on 2 this post and realized you are talking about MOTORCYCLE horsepower!
Mercy sakes alive!
Wouldn't send my kid out on that rocket!
If he was talking about a motorcycle that had 622 hp it better go faster than 179.
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Howdy All, :-D
Since you mentioned 179mph and the initial thread was about dyno speeds. :?
I have gone 179mph on a bike and although it was relatively easy. :| It scared the stuffing out of me and I don't spook too easily. :-P
Of course, this was only on the chassis dyno and I am well convinced that one of those streaks would skin me up in a heartbeat as my skill level would be overcome before the thing got out of second gear! :roll:
You guys on the crotch rockets continue to amaze me and my hat is off to you all.
Regards to All,
HB2 (smiling)
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I would like the spreadsheet you can send it to stites@cox.net
Aaron, since you are talking about gearing I have a question? I am currently geared at 2.0, 24 front 48 rear and the stock gearing is 1.9 or 1.8 if I recall correctly. I thought 2.0 would be a good place to start to see if the bike can pull the gearing rather than be to optimistic and work my way down in teeth on the rear sprocket. Your thoughts??
Scott
ME TOO PLEASE :wink:! kazracing@cox.net
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BTW the chassis dyno showed we made 518rear wheel hp (505 uncorrected) an 462 tq (really flat with the tight converter) through a TH400. This was on a Mustang Chassis dyno at Delk Performance in TN. There was no CD info in their book for a 59 Jag so we picked a generic mid 80's Mercedes E type as a relative comparison to input as well as the vehicle's weight of 2650 plus driver. We pulled the car up to 190 on the dyno and it was still creeping up but slowly. It showed peak hp at 180mph - coincindentally basically our exit speed at Maxton - Or maybe not so coincidentally.
Anyhow, it showed us that we need to make just over 500 uncorrected hp at the wheels to run 180 at Maxton in that car. AT least that is A data point.
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The amout you need to gear short can bee figured by using the tractive effort formula qnd you eng. dyno. sheets:
TQ. eng torque
FDR Final drive rato--Trans Ratio X Rear Axle Ratio
TC you use one of you tires as a value of 1.00-- then use the a 1.1 for a tire that is 10% smaller and .90 for one that is 10% larger
FDR X TC==Overall Gear Ratio or Total Gearing---or what ever you wish to call it.
Tq. X FDR X (TC)= tractive effort
you can put this on a spread sheet for you eng speeds and figure your speeds in each gear "Your Dream Chart" as well as "what is being put to the ground" no driveline losses figured but they should remain almost the same. This will tell you when to change the Overall Gearearing for better speed---My lakester is at the cross over point. Right now, It will probabally run the same speed with the 2.28 or the 2.14s---I will be cleaning up the body this winter and installing the 2.14s-- peak Hp is about 6800 and I am turning 7050+ for the last mile.
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The amout you need to gear short can bee figured by using the tractive effort formula qnd you eng. dyno. sheets:
TQ. eng torque
FDR Final drive rato--Trans Ratio X Rear Axle Ratio
TC you use one of you tires as a value of 1.00-- then use the a 1.1 for a tire that is 10% smaller and .90 for one that is 10% larger
FDR X TC==Overall Gear Ratio or Total Gearing---or what ever you wish to call it.
Tq. X FDR X (TC)= tractive effort
you can put this on a spread sheet for you eng speeds and figure your speeds in each gear "Your Dream Chart" as well as "what is being put to the ground" no driveline losses figured but they should remain almost the same. This will tell you when to change the Overall Gearearing for better speed---My lakester is at the cross over point. Right now, It will probabally run the same speed with the 2.28 or the 2.14s---I will be cleaning up the body this winter and installing the 2.14s-- peak Hp is about 6800 and I am turning 7050+ for the last mile.
Hey Sparky you tried to explain this to me once and how it relates to racing, but I can't get my head around it. When I do a google search I come up with how it applies to trains:
Tractive Effort (abbr. TE) is the pulling force exerted, normally by a locomotive, though the term could also be used for anything else that pulls a load. It is normally understood to be the actual force on the locomotive's drawbar or rear coupler. When a bare figure for tractive effort is quoted without a speed qualification, this is normally for starting tractive effort, i.e. at a dead start with the wheels not turning...................
..........................For a locomotive to accelerate from a stationary position, it must apply a force to overcome the inertia of the train, along with the frictional forces in the form of mechanical friction, and wind resistance as the train accelerates. In order for this to occur a particularly high tractive effort is required, usually the maximum tractive effort of the engine is applied. This means that the engine works to produce the highest possible force that it can exert onto the wheels to cause movement or motion. Few engines can maintain work at the maximum tractive effort for very long, but neither is it usually necessary for an engine to do this. Once the train is running at a constant velocity the train no longer needs to overcome its inertia to remain at the same velocity, and hence must only provide power to compensate for frictional forces. This leads to one potential upper limit on the speed a locomotive can haul a train at, once the force due to wind resistance becomes greater than the tractive effort the locomotive can supply (fluid drag increases with the square of velocity), the locomotive cannot accelerate the train anymore (in reality the situation is more complicated than this due to a number of mechanical considerations).
The last part sounds a lot like our problems:
"upper limit on the speed....once the force due to wind resistance becomes greater than the tractive effort the locomotive can supply (fluid drag increases with the square of velocity), the locomotive cannot accelerate the train anymore"
So how do I use your formula and this concept to figure my gearing, HP needs, traction needs???
Thanks and maybe you will have to explain it when you get to my place later this month,
Sum
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Sum, I think you are trying to make it too complicated---all we are trying to do is compare the force to the ground at a given speed with different gear and tire posibilites; and we need to go with combo that produces the largest TE.
As the torque falls off past peak HP, are we better going with a 5% lower gear because at 5 % more revs. the eng torque is only down 3% so we have a net gain of 2% TE.
.. I my case I am past peak TQ to the point that my TQ cure is really beqining to fall off pretty fast; I am going to proabably better off changing the gear which will move me 7% back down the TQ cure so that the net result is that I am putt more TE to the ground---when we get against the Aero Wall --ever little bit helps..depending how much I can clean the car up we are only talking 3-5 MPH in my quess.
Bototm line TE just lets you compare at a selected speed either what you have already done or what you are shooting for---does nothing to tell you how much you are going to need!!!!!!!!
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As the torque falls off past peak HP, are we better going with a 5% lower gear because at 5 % more revs. the eng torque is only down 3% so we have a net gain of 2% TE.
The scenario you describe is a physical impossibility. You cannot be at a 5% higher rpm than your hp peak, with 3% less engine torque, because that combination would result in more horsepower, not less.
Horsepower is just torque times rpm. Raise rpm by 5% and reduce torque by 3% and you have more horsepower. 1.05 x .97 = 1.0185. So to be precise, your scenario has 1.85% more horsepower.
The maximum rear wheel torque at any given rear wheel speed occurs at the power peak. You can't gain rear wheel torque by revving past the power peak and then gearing deeper to get back to the same speed. You'll have less rear wheel torque if you do that.
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You can't gain rear wheel torque by revving past the power peak and then gearing deeper to get back to the same speed. You'll have less rear wheel torque if you do that.
But, as you pointed out earlier, gearing it a little short makes it a lot easier to accelerate to around your peak speed, and is way better than being geared too high.
By the way, in this month's Street Rodder, their tech guru, Ron Ceridono, says "a car will accelerate the hardest when the engine is at its torque peak". (I'm not making this up.)
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ASW---I am refering to the thrust at the rear wheel--by using the higher number ratio you get a larger number at the wheel because of the larger gear multiplier--- vs gearing for peak torgue---you have a bigger number to the trans and rear axle---but loose it by the smaller rear gear multiplier---you can multiply torque---but not HP---bottom line---its all about what you can put to the ground and hook up.---
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Tortise---the statemet you quote applies to the same overall ratio---and probabaly does not necessarily apply to LSR cars that run way over 2---because of wind resistance.
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ASW---I am refering to the thrust at the rear wheel--
So am I.
by using the higher number ratio you get a larger number at the wheel because of the larger gear multiplier---
Of course. Gearing multiplies torque. The more rpm you turn at the engine, the more gear reduction you can apply for a given rear wheel rpm, hence the more you can multiply the torque. I understand all that.
vs gearing for peak torgue---you have a bigger number to the trans and rear axle---but loose it by the smaller rear gear multiplier---you can multiply torque---but not HP---bottom line---its all about what you can put to the ground and hook up.---
Here's where you lose me.
Torque and rpm are interchangeable entities. Gear taller, you give up rear wheel torque but gain rear wheel rpm, and you do so by equal factors. Gear shorter, you gain rear wheel torque but you give up rear wheel rpm, and again, you do so by equal factors.
Because they're totally interchangeable, each of these two items, torque and rpm, are weighted the same in the horsepower formula. Horsepower is literally torque times rpm. They make an equal contribution to the performance because they're totally interchangeable. And they make absolutely equivalent contributions to the performance. So horsepower describes performance.
When you're at your power peak rpm, that's the point where the total combination of torque and rpm are at their greatest. This notion that you can run your motor past it's power peak rpm, and then gear shorter such that you get back to the same rear wheel rpm you had at the power peak, and end up with a higher combination of torque and rpm at the wheel than you had before, is patently false. If you could do that, you would be raising horsepower with gearing, because your total combination of torque & rpm would be higher on the downstream side of the gearing. Gearing can't raise horsepower, it can only change the mixture of torque and rpm that makes up the horsepower.
Bottom line, the power peak rpm of the engine represents the point where the combination of engine rpm and torque are at their highest. You cannot artificially create a higher combination by revving higher and gearing deeper.
If you're still convinced there's something to be gained by revving past the power peak and gearing deeper, give me an example. I can show mathematically that it doesn't work.
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But, as you pointed out earlier, gearing it a little short makes it a lot easier to accelerate to around your peak speed, and is way better than being geared too high.
Yep.
By the way, in this month's Street Rodder, their tech guru, Ron Ceridono, says "a car will accelerate the hardest when the engine is at its torque peak". (I'm not making this up.)
Wow.
Whenever someone has tried to make that argument to me, they generally argue that the car "accelerates hardest in any given gear at it's torque peak".
Which of course, is true, but misses the point. In the real world, we're not constrained to staying the gear we're already in. We have the option of downshifting and putting the motor at it's power peak instead. And when you do that, you accelerate even harder.
A very simple example that most people can visualize is to get two identical vehicles moving side by side, both at their torque peak. On one of them, floor it. On the other one, downshift and then floor it. It won't even be close. The downshifted one will run away.
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If you have a dyno sheet---plot your own---I have done it with two different engs the first was more head constrainted than the current one so it had a flatter torqe curve---this new eng has more HP---but falls off much faster---on the spread sheet I am at he cross over point---next year I will know for sure if it works out in the real world---the first eng I was better of staying with the lower gear. WE are not talking about big numbers---15-30' # difference on a BBC. I shift at 8 and am pull 70-71 out the back nowI hope to work it out so that I can pull the same with more gear.
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aws---Your own example prove my point---I am just downshifting the rear gear or tires-instead of the Trans---to change the FDR
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Sum, RPM vs. TQ What you say is true about them being interchangable---but when you put the non linear curve of the eng. Torque x by the overall ratios vs linear speed --- get some crossover one combo vs another combo at the same speed. I you can pull both and one is falling off faster than the other-- one should try the combo with the most Tractive effort to try to accelerate.
But as The Mayor has pointed out---its still VERY Important--- the 2 1/4 ET---the 60' of LSR--it still is a drag race-- it just happens to be for 4 miles with a one mile trap.
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Sum, RPM vs. TQ What you say is true about them being interchangable---but..............................
I don't think I was the one who said that.
One thing I'm not familiar with that was said above was:
the power peak rpm of the engine represents the point where the combination of engine rpm and torque are at their highest.
I've never seen combining the engine rpm and torque together before and would like to know more about that concept.
Sparky I think I'm beginning to understand a little more where you are headed with with this. Since in the HP/Torque formula we have 5280 that is the point where they are the same.
(http://www.compcams.com/information/Products/Camshafts/DynoSheets/xr276.JPG)
Above and below that point and even through that point they typically are on different curves (angles) and thus are rising or going down at different percentages. I think you are trying for gains by working in the area where the torque begins to fall off rapidly in the area where the HP is peaking or just past peak.
I made a spreadsheet today, but didn't have time to play with it and want to plug in some dyno numbers to see if I can understand this better. So with the tractive effort are you trying to maximise thrust at what you hope will be your top speed??
c ya,
Sum
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Sparky I think I'm beginning to understand a little more where you are headed with with this. Since in the HP/Torque formula we have 5280 that is the point where they are the same.
Sumner,
I am pretty sure that number is 5252 rpm. Hp and tq always cross at 5252 and the forumla to figure out Hp is........
torque x rpm/5252
The constant 5252 is derived from James Watt's 17th Century definition of horsepower. One horsepower is equivalent to performing 33,000 pound-feet of work in one minute. If 33,000 pound-feet per minute is divided by pi (6.2832) multiplied by 2, the result is 5252.1008 rounded to 5252.
An example using the graph you posted
That motor makes its peak torque of 425 at 3750 rpm
torque x rpm / 5252
425 x 3750 = 1593750 / 5252 = 303.45 Hp at that same rpm
Scott
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Sum,
I am ALWAYS working to try to figure a combination of tire size, Overall tio that will geve me a better TE AT OR ABOVE my previous best speed---in my case I have run 276 with an exit of 279 with MT 30s and 2.28 --- around 282 my TE with 2.28s and 2.14s are the same---below 282 I have a better TE with the 2.28--
282 and above I have a better TE with 2.14s which move me back down the TQ curve by 7% which has a TQ value of greater than 7 % at 282--therfore I will have more thrust at the rear wheels with 2.14 and app. 6500 than I would have with 2.28 and 7300.--- I may be giving up some accleration at lower speeds --(MAYBE)---I have had an issue of hooking it up so far with the tracks that we have had--
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If for example: I could get to 300 then I will be better of going to my MT 26" and my OD which will give me a FDR of 1.77---I doubt without a snoot full of NO2---that ain't gona happen!!!!!!!!!!!!! except in my dreams---- LOL
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Sum, I am dealing with a similar crossover---but mine is the TE of the different FDR and tire combinations:
ie overall ratios multiplied by the eng torque at 100 rpm intervals on the spread sheets!!!!!!!!!!
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Sum, Scott is right. There is a simple mathematical relationship between torque, rpm, and horsepower. Horsepower is torque times rpm. Yeah, you divide it by 5252 to scale the result down to match what Watt's horse could do, but conceptually, you can think of horsepower as torque times rpm, i.e. the combination of the two. As I previously showed, torque and rpm are interchangeable entities that make equal contributions to performance, therefore they're weighted equally in the horsepower formula.
The 5252 scaling factor comes from the fact that Watt's horse could lift 1lb 33,000 feet in 1 minute. Well, 1 lb-ft of torque is defined as 1 lb of rotational force at a 1 foot radius. 33,000 feet is how far that 1 foot radius travels if you make 5252 rotations (r x 2 x pi x 5252 = 33,000). So 5252 rotations moves that 1 foot radius point 33,000 feet. If it has 1 lb of force on it and it does those 5252 rotations in 1 minute, it's doing exactly what Watt's horse could do and we call it a horsepower.
In any event, the important point is that the power peak rpm is always where the combination of torque and rpm are at their greatest. If you go to a higher rpm than the power peak, the torque will drop more than the rpm rises, hence you have less horsepower.
My point here is that it's physically impossible to rev the motor past it's power peak, and then with gearing, end up with some higher combination of torque and rpm. The power peak is where you want to be when the wall of air stops your acceleration, because that's where you're getting the most out of the motor. You can't get more with gearing tricks.
That's what you want to do to maximize rear wheel torque at a given rear wheel rpm. The difference between that and force at the contact patch is a function of the radius of the tire. But it scales linearly, there's no free lunch. A shorter tire reduces ground speed by the same factor that it increases contact patch force, and vice-versa.
That said, I think I misunderstood what Sparky is doing. From his later post, it seems he's really effectively adding resolution to his gearing selection by considering tire size in conjunction with axle ratios and power available on each side of his power peak. Basically, since he doesn't have infinite resolution in gear selection, there's a range of rpm he's forced to work over, he can't always gear right for his power peak. His available axle ratios and tire sizes define that resolution for him. Anytime he's not right on his power peak, he's going to have less than optimum rear wheel torque, but that may be outweighed by being able to get closer to the wall, which of course is the name of the game.
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Sparky I can see where you are going with this ( I just did a little spreadsheet and pasted in some dyno numbers) but please be sure to mention that you can't blindly enter in a new torque value at the new rpm without going back to determine the correct gear ratio (or tire size) that would appropriately lead to the new and improved TE.
In other words looking at a dyno sheet here that a 434 was making 511.3ft lbs at 6000 and had a TE of 1308.9 with a 2.56 gear, 1:1 trans and a 1.0 tire ratio. We now look at 6200rpm and the engine is now making 494.7ft-lbs. Is that better or worse? Keeping all things equal it is worse - 1266.4. Yeah we can arbitrarily enter in a new gear ratio or tire ratio to inflate that number but we have to keep in mind the our new gear ratio/tire ratio MUST have the engine running at 6200rpm now in the real world. If we don't keep the reality of an rpm change in mind then the numbers don't mean anything. OK so now we have to reverse engineer the formula to determine the RPM of the engine. I would recommend altering your formula to use an actual tire size instead of a multiplier to more easily determine this. And to do that we need to take into account MPH.
Please take this assignment home and bring it back to class tomorrow with your proof.
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Sum, Scott is right. There is a simple mathematical relationship between torque, rpm, and horsepower. Horsepower is torque times rpm. Yeah, you divide it by 5252 to scale the result down to match what Watt's horse could do, but conceptually, you can think of horsepower as torque times rpm, i.e. the combination of the two. As I previously showed, torque and rpm are interchangeable entities that make equal contributions to performance, therefore they're weighted equally in the horsepower formula..............................
Ok I put the 5280 down because I was too lazy to go look up the 5252. Hey I was only off by 28 and that ain't bad when you are 63 and at least I know how long a mile is (http://www.cybergifs.com/faces/glasses.gif) , and I've seen the HP/Torque formula tons of times. Glen and I got to ride Watt's horse when we were younger in fact.
..................In any event, the important point is that the power peak rpm is always where the combination of torque and rpm are at their greatest. If you go to a higher rpm than the power peak, the torque will drop more than the rpm rises, hence you have less horsepower.......................
Ok so let's get down to the above statement. Here are some dyno numbers.
RPM HP Torque T + RPM T X RPM HP X RPM
2000 138 363 2363 726000 276000
2500 181 379 2879 947500 452500
3000 227 397 3397 1191000 681000
3500 278 418 3918 1463000 973000
4000 326 428 4428 1712000 1304000
4500 368 430 4930 1935000 1656000
5000 391 411 5411 2055000 1955000
5500 393 375 5875 2062500 2161500
6000 372 326 6326 1956000 2232000
6500 337 272 6772 1768000 2190500
You said to combine the rpm and torque and I wasn't sure from another post if you meant add them together or multiply them, so I did both. The adding together (4th column) didn't seem to work as the resulting number always got higher even when the torque was falling off at 6500. The multiplying method (5th column) seems to make sense and at 5500 you have the highest number and that is also the HP peak. Is this the method you use??
If I was gearing I would have probably picked about the same point based on the HP peaking there. The 6th column show HP X RPM and if you used that column 6000 rpm would be the peak and that is probably where I would shift on the drag strip, on the salt in the first couple gears for a high HP car short shifting seems to help as you can never get full throttle in these gears anyway.
Looking at all of this I think we are all coming to about the same conclusions just from slightly different perspectives and until you have some real world results on the salt or dirt it is hard to fine tune the gearing to achieve the maximum potential of the vehicle at any set HP/Torque level. Before you run you are doing a lot of guessing with the aero drag.
I'll add your torque/rpm to my data that I look at and also Sparky's Tractive Effort.
Thanks guys,
Sum
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The 6th column show HP X RPM and if you used that column 6000 rpm would be the peak and that is probably where I would shift on the drag strip
At the drags you want to shift at above the HP peak if the redline allows, and cross the line at above the peak, to keep the RPM range in each gear more or less centered around the HP peak.
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The 6th column show HP X RPM and if you used that column 6000 rpm would be the peak and that is probably where I would shift on the drag strip
At the drags you want to shift at above the HP peak if the redline allows, and cross the line at above the peak, to keep the RPM range in each gear more or less centered around the HP peak.
6000 was above the peak HP in the example. The peak was at 5500, but the HP X RPM peak was at 6000. I found that interesting as that peak goes along with the commonly accepted shift point that you mentioned.
c ya,
Sum
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6000 was above the peak HP in the example. The peak was at 5500, but the HP X RPM peak was at 6000. I found that interesting as that peak goes along with the commonly accepted shift point that you mentioned.
c ya,
Sum
I should have read you more closely. Calculating the HPxRPM peak is a prosecutable case of spreadsheet abuse, I fear. The amount above peak HP to shift at is a function of how close the gear spread is.
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Glen and I got to ride Watt's horse when we were younger in fact.
[/quote]
LMAO!!!
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Glen and I got to ride Watt's horse when we were younger in fact.
LMAO!!!
[/quote]
HORSES TO LET AND DYNOS VERIFIED
INQUIRE WITHIN
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bla..bla...bla...
Sum put your calculator down.... go out in the garage and pick up a wrench... you got 289 days till speed week... and less if ya wanna take your lakester to elmo for a pre-speedweek shake down
love ya
kent
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Tortoise --gets IT as I understand IT --we are a 4 mile drag with a 1 mile trap---almost the same---I am just using TE --trying to figure out how much to under gear.
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The 6th column show HP X RPM and if you used that column 6000 rpm would be the peak and that is probably where I would shift on the drag strip
At the drags you want to shift at above the HP peak if the redline allows, and cross the line at above the peak, to keep the RPM range in each gear more or less centered around the HP peak.
6000 was above the peak HP in the example. The peak was at 5500, but the HP X RPM peak was at 6000. I found that interesting as that peak goes along with the commonly accepted shift point that you mentioned.
c ya,
Sum
I should have read you more closely. Calculating the HPxRPM peak is a prosecutable case of spreadsheet abuse, I fear. The amount above peak HP to shift at is a function of how close the gear spread is.
I'm confused, which is it?? When drag racing shift above the HP peak or not?? You say "shift at above the HP peak" The HP peak on the dyno run I posted was at 5500 and the point where the HP X RPM produced the highest number was 6000 RPM. Is this not above the HP peak??
Now I didn't say use the HP X RPM method as awsracing never said to use it, I just observed that it indicates what your are suggesting "shift at HP peak".
I agree that the gear spread is critical and why I made the spread sheets on my site that show the rpm drop on gear shifts at any rpm:
http://purplesagetradingpost.com/sumner/bvillecar/bville-spreadsheet-index.html
c ya,
Sum
P.S. Kent I've been working on the car since your post and I'm going back out there now :-D .
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Now I didn't say use the HP X RPM method as awsracing never said to use it . . .
All I was saying is that the HPxRPM number ain't got nothin' to do with nothin', and wondered what it was there for. We've run out of stuff to argue about.
c back atcha,
tortoise
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"HPxRPM number ain't got nothin' to do with nothin"
True - HP × RPM is TQ × RPM × RPM, doesn't exist.
The "combination" is TQ × RPM, not TQ + RPM.
"The amount above peak HP to shift at is a function of how close the gear spread is."
And how flat the torque curve is, and how much change in resistance occurs with higher speed.
If torque falls off quickly after peak power (common in highly developed resonant systems tuned for peak power) shifting at or immediately after peak power produces better results.
If torque is high (but decaying) after the peak you can use a higher RPM figure than peak.
If torque below the peak is poor, even more RPM above peak is needed.
From a practical perspective, near maximum speed the engine must recover to an RPM level at which the power is almost as high, or the car will slow down ("the damned thing goes faster in 3rd than 4th"). Really fast cars, or peaky engines need a very close shift into high gear, and can use more "progession" (the spread in RPM drops through the gears, with the same 1st gear), which would have little value in a drag race.
An example of high progression, #1:
1st 2.50:1
2nd 1.55:1 recovers 62% from the 1-2 shift
3rd 1.10:1 recovers 71%
4th 1.00:1 recovers 91%
This is better than less progression with the same 1st and 4th, #2:
1st 2.50:1
2nd 1.65:1 recovers 66% from the 1-2 shift
3rd 1.25:1 recovers 76%
4th 1.00:1 recovers 80%
#1 has less total power under the RPM curve than #2, but is still has higher top speed, because the critical 3-4 shift is the only one that occurs when the car is aero-limited rather than traction-limited.
Why not just use all close ratios, like #3?
1st 2.00:1
2nd 1.40:1 recovers 70% from the 1-2 shift
3rd 1.10:1 recovers 79%
4th 1.00:1 recovers 91%
For some transmissions it doesn't exist, or $$$. For some, the much lower numerical 1st gear (i.e., 2.00:1 instead of 2.5:1) makes launching the car very difficult (clutch slip, converter heat problem, plug fouling, spark curve), and the engine must develop more power at low speed to overcome this.
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After many months studying torque compensated computer rotational dynamic simulations and gathering empirical data along with advanced scientific evaluation I have derived the following information which, because of my unselfishness and devotion to land speed racing I am willing to share with you all.
Put a gear on it that makes it go the fastest.
“Theoretical records are set by theoretical vehicles”
Quote by Denis Manning
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"Yup , just like they said and said and said, a dyno might float yer boat, but it won't teach you how to sail." (me)
I wonder where Denis heard that ? :wink:
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and it takes the 2nd impound trip to make it non theory--- as the AA GL boys found out this year at SW ---5 one way trips 4 by the current record holder and 1 by me ---"just a tryin to "GET'r Done"
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Panic,
My FDRs are
1. 4.01
OD 3.13
3. 2.28 GM rear axel
OD 1.77
1. 3.77
OD 2.94
3. 2.14 GM rear axel
4. 1.67
As you can imagine I can only run the OD in direct with 26" tires as opposed to the 30"s---which I haven't tried yet---but will soon.
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Sparky. I built a spreadsheet last night with inputs for tire size, rear end gear ratio, converter slippage (for autotragic cars), TE constant tire size with output info for mph based on rpm at up to 6 different gears as well as TE outputs at the rpm mentioned. I'll bring home some dyno sheets tonight to plug in some real data to how how this whole deal works with the TE info.
I had an old spreadsheet that was similar that I used to show torque drop for shifting rpm but it's lost in the puter somewhere for now.
I'm sure most folks here have similar spreadsheets but if anyone wants an excel copy let me know.
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Rev, I am intrested is seeing how you have set up your converter slip---I presently don't use a converter--but will when I am able to go to Texas and or Maxton---with what you have done in the past you may already have done most of this---remember what I was trying to accomplish---what combo that I can max out for a given speed---If I get to run at EL M this weekend I will be using the 30" tires with my 3.13 FDR comb to try to run 220+---I will be turning 8M +---way past my peak eng power---but will maximise my TE!!!!!!!
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Rev. Dont for get---Its eng(TQ) X (TR)Trans. Ratio X (RA) Rear Axle ratio X (T) tire
I use a value of 1 for my most commonly used tire my 30"---and .00 Adjust for any % size difference-- I am calculatin rear end thrust or Tractive effort---what is pulling the car forward --asuming it is hooking up
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Dyno. A scientific device for determining engine tune up.
Bonneville. A scientific device for determining gearing. And lowering your expectations.
Determining HP on different engine setups is what a dyno is for.
Gearing is a SWAG at best. You don't know what the aero figures are going to be. Two identical vehicles with different aero are going to require different gears.
Air flow through the intake on the salt can be far different. And never in a good way.
The dyno doesn't rattle your teeth like the salt does. Has much better traction than the salt too.
As far as those 5 mile pulls at the salt, you should be doing them on the dyno. If it won't do it there, no sense in putting it on the trailer.
Having a setup that allows fast gear changes beats a spreadsheet every time.
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Dean--that quick change EATs power---with different tire sizes and GM ratios with a GV OD---I can get awfully close---I need to work on AERO now!!!!!!!!
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Dean--that quick change EATs power---with different tire sizes and GM ratios with a GV OD---I can get awfully close---I need to work on AERO now!!!!!!!!
Spark, Dean and all the bike guys don't have the power eating gear change problem... sprockets, fronts and rears, lots of changes, works the same for the bike motors in cars....
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great point!!!!!
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There's just the reason I needed to go back to work on my AA / Thomas Flyer roadster!