Drivetrain Power Loss

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TFF

Well-Known Member
I would surmise that most of these gearbox’s are closer to 70s numbers than 2010s when it comes to losses. Most do not have access to the quality that a big auto maker has.

DangerZone

Well-Known Member
HBA Supporter
No, I postulated a robust 90 degree bevel gear set that would likely be durable, calculated the pitch line forces for 100 hp and 7500 rpm, then calculated the likely losses by the gear mesh and bearings on the shafts. I did the calcs using a rather conservative 0.005 coefficient of drag for steel rolling along on steel. That is how gears and roller bearings are modeled by folks who design these systems, and usually the folks in automatic transmissions assume even smaller drag, because, quite frankly, automotive transmission gears and bearings do better than this. The method can be expanded to cover real gear sets with known dimensions, but it is common for these gear sets to do modest fractions of 1% loss. Oh, and this is way more info on modeling than that the pack of... info... you have been supplying.
Ok, I see. So, you basically calculated only the rear bevel gear losses without the losses from bearings, the swingarm deflection under load, the differential, the dampener, the rubber boots, the oil/grease, the small shaft clutch, the two U joints which are also part of the shaft drive? This makes sense and allows me to understand why you got only 1% losses. I can't afford that, I usually have to use the school engineering method of calculating every single component loss and then calculate all losses of the whole system together.

I feel it is necessary to point out that the example run through in the article is illustrative of the method, but is also quite conservative and does not make any attempts to calculate the friction, then torque lost, then energy losses of real systems. As I mentioned in an earlier post, in the automatic tranny world, we are accustomed to whole gearbox efficiencies around 97%, and that is with around half of the clutches open and dragging as well as spinning quad compound planetary, transfer, and final drive gears. The example used provided no basis shown for each of the numbers. The drag in individual gearsets was quite high compared to anything used in modern light truck or car powertrains since at least 1990 MY, where my professional involvement in these machines stretched started.
Yeah, it seems the methods are different. Ok, so how would you calculate the losses from for example a Honda Accord 2.0i from 2007 with 155HP at the crankshaft, one automatic and the other manual, what methods would you use? They both have the same engine, only the drivetrain is different. This was a very common Japanese car in EUrope, and the manufacturer had to comply to EUropean homologation rules.

Says who? Certification in the US specifies the start conditions, dwell before driving, then the speed vs time trace for city and highway cycles. On the test, measured is actual speed vs time over the cycle, fuel used and emissions produced. The rest of the cert is that the evaporative emissions control system works, and the whole thing has to be demonstrated that it can be expected to go 100,000 miles without going smoggy. Once the vehicles are being sold, the fleet is allowed a very small number of emissions related failures before a target mileage number or a recall to fix the systems is triggered. The Euro and Asian cert process were modeled on the EPA processes, but with a little more rigor in finding a test cycle that reflects actual usage a little more than the EPA cycle did. In 1970, the data collection and sorting was less sophisticated than it is now. Story is the EPA lab instrumented a car, had ten employees drive them home and back, and picked the secretary's trace as looking closest to the middle of their range. That cycle is still used 50 years later.
It is different in the EU. Many EU countries charge taxes by the engine kilowatts, so this is quite strict. Before kilowatts it used to be horsepower until 2010, and then the official measurement became the kilowatt while horsepower was only noted on the side. A kilowatt is (in countries which accept the DIN standard) the exact power necessary to lift 75kg (hence 75 kiloponds) 1 meter high in the time of one second. It is pretty exact, just like gravity. The problem was that many Japanese and American cars did not pass homologation because they were out of EU technical specs or even had different horsepower measuremnt. Some great American cars still can't be imported to the EU because their specs and reality do not match when they are put on a dyno, or some other specs are different and they do not comply to EU rules.

Again, says who? You are assuming way more government oversight than is actual. The emissions and fuel use are what are tracked. If the car is a dog, they do not care. Oh, there are lots of safety checks in FMVSS, but measured power? Nope. And most of it is self certified too.
That would depend from country to country. The EU has many countries and many apply strict rules, but some really tend to tolerate self certification and bend the rules. Some really measure power. Some don't even have facilities which can measure power, but they require them by law. It can be a mess regulation wise sometimes. I find the American approach much better, I doubt paying taxes according to kilowatts/horsepower would seem normal in USA. Oh, and the pollution hysteria in the last decade? On top of kilowatt taxes, they also charge pollution taxes. Like, a V8 is breathing too much hot air into the atmosphere. I installed LPG in most cars I owned, but they also charge a special tax for that justifying this by saying that we drive cheaper on LPG. Typical double standards, it does not matter if my car is polluting less, it is bugging the administration that I drive cheaper and they get less money from overpriced fuel. Ever wondered why fuel is so expensive in EU compared to USA?

Again, I suspect exaggerated power output claims, poor calibration of dynos, and very likely, a fair amount of slip from the drive wheel to the roller - typical automotive tires peak in load carrying capacity at around 7% slip. A 100% powertrain can not do better than the speed ratio between tire and rollers. Chassis dynos are great for durability tests though.
Ok, so are you saying that you do not verify your drivetrain losses calculation on a dyno after doing all the math? How will you know then that your losses and efficiency are exact?

DangerZone

Well-Known Member
HBA Supporter
I raced 100cc bikes a lot as a kid. The larger the difference in gear sizes made a large impact on drive line loss. Just like they tellyou in Mech engineering,
For example a 100cc Hodaka in 4 th gear ,driving a certain , rear gear sprocket, made 12hp. at the wheel .
In 5th gear, driving an larger sprocket for the same IPR( inches per revolution) the bike put down 11HP at the same wheel and speed.
Keeping the gears as close to the same size as possible will produce the most power .
Keeping the gears straight cut will also help out put by reducing side loads and slipping loads. Less heat = more power out.
FWIW, my chassis dyno allows me to measure coast down drag. Some gear sets are 22 HP on the FWD VW, that I race and tune. That is my upper limit. I rebuild them with loose fit bearings etc at 22HP.
On the same note, i found that running in 3 rd gear gives more power then 5 th gear . similar to the Hodaka 30 yrs agao. I set up the best cars to spend as much time in 3 rd gear as possible now. The 3rd gear sizes are large and closest to the same size, as is the final drive gear set.
We had similar experience when changing sprockets on 1000cc bikes, specially when changing front sprockets to a smaller tooth count for better acceleration and handling. Even though some would expect higher revs and better runs, the efficiency went down in higher gears. The manufacturers had obviously done a great job at testing, and found the best or most optimized front and rear sprocket ratio.

Geraldc

Well-Known Member
I have read the to and fro comments and think the main difference that is not taken into account by dangerzone is tyre loss on a dyno.
Axle mounted dynos prove this.

Dan Thomas

Well-Known Member
Ok, I see. So, you basically calculated only the rear bevel gear losses without the losses from bearings, the swingarm deflection under load, the differential, the dampener, the rubber boots, the oil/grease, the small shaft clutch, the two U joints which are also part of the shaft drive? This makes sense and allows me to understand why you got only 1% losses. I can't afford that, I usually have to use the school engineering method of calculating every single component loss and then calculate all losses of the whole system together.
Just how much power do you thing swingarm deflection could absorb? What differential would you have in a bike, and in any case how does a differential lose power if both shafts are being turned at the same rate so that the spider gears are not turning? How does a clutch lose power when it isn't slipping? How do U-joints lose power if their shafts are lined up? How do rubber boots absorb power when they're not even turning?

I think you're well out in left field here and don't understand that HP is torque times RPM, and to lose much HP through efficiency losses you have to have some serious heat-generating friction somewhere. Swingarm deflection, for instance, isn't going to create enough frictional heat to absorb even half a watt. U-joints lose nearly nothing, even if the shafts are angled at 10° . Their needle bearings are pretty efficient. Rubber boots absorbing HP? Come on. On and on it goes. Like has been said multiple times here, you'd be coking the oil in that thing, or melting it down, if it was it was losing 18 HP. That's serious kilowattage there. Physics and math, every time.

Aesquire

Well-Known Member
Beat me to it.

The number 1 reason to convert a shaft drive motorcycle to chain drive is to have options for gearing.

shaft drives come from the factory with one gear choice. You don't order gear sets to change that since, simply, they don't make them.

Weight, style, and a desire for a different length swingarm are other reasons but by far #1 is to have changeable gearing. Shaft drive bikes have torque reaction quirks that bother some, but mostly their owners just adapt. I had a Yamaha XS 750, and personally liked that I could use the torque reaction to my advantage.

My current motorcycle, a '01 Buell Cyclone uses a belt drive. Strong, reliable, quiet, and no oil spatter. However, my gearing options ( which are a rare thing on belt drive bikes ) are limited to "U.S. & Euro" gearing. 2 front & 2 rear sprockets I can mix & match. That's 4 possible ratios. The engine has a rev limiter. So top speed on my current sprocket set is 135 mph. My "tallest" choice gets me 155 mph.

Thus it is common for racers to convert to chain drive to have far more gearing choices to match the track & riding style. For example, if I wanted to set land speed records, I would gear for my anticipated top speed. If I wanted to race at Watkins Glen, I'd gear for best compromise between acceleration and top speed.

Much like the difference between a climb & cruise propeller.

The speed differences between a shaft drive and chain drive bike of The Same Brand, Displacement, etc. As mentioned above, is simple. Not the same purpose machine. The cylinders & head castings may be the same, but cams, porting, and gearing are chosen for the purpose. Touring with low maintenance and no oil on your clothing, or Performance with gearing choices and typically More Power and a DIFFERENT torque curve.

modern bikes can have the power and handling to be quite versatile. But it's a set of compromises.

just like a RV-7 & a Pitts S-2 are capable of acrobatics and flying to the pancake breakfast.

Aesquire

Well-Known Member
Btw, on 2 wheel motorcycles with shaft drive, there is NO differential.

as noted by multiple people above, the difference in CARS between crank and chassis Dyno PUBLISHED BRAGGING, is due to both massive losses in the tire/roller interface, and because the dyno figures Marketing uses are without any accessories needed to actually operate. ( usually. Sometimes there is an alternator, and often the stock water pump is used. )

And of course, you only publish power numbers obtained on an engine at the perfect temperature for oil drag and intake manifold. Corrected for STP.

There's a show on Motor Trend called "Engine Masters" where they test various parts etc. on a calibrated dyno back to back. You can make several more ponies with the right amount of oil in the pan.

Pilot-34

Well-Known Member
My wife says I need to withdraw from this thread she’s tired of washing my clothes.

wsimpso1

Super Moderator
Staff member
Log Member
Ok, I see. So, you basically calculated only the rear bevel gear losses without the losses from bearings, the swingarm deflection under load, the differential, the dampener, the rubber boots, the oil/grease, the small shaft clutch, the two U joints which are also part of the shaft drive? This makes sense and allows me to understand why you got only 1% losses. I can't afford that, I usually have to use the school engineering method of calculating every single component loss and then calculate all losses of the whole system together.
No you do not see. Specifically:
• I discussed the bearings on the 3/4" diameter shaft, their loads, the friction forces, the friction torque and then the power used turning them. they were part of the total;
• Swing arms major deflection occurs once when driveshaft torque is increased, and then unwinds when driveshaft torque decreases. So once per throttle cycle or gear change if power is release during the gear change, less when shifted without the clutch. At firing order there should be very low deflections, on the order of thousandths. So if we have 840 in-lb torque and say 0.005" firing order deflection, that is 840 in-lb * 0.005 * 125/s = 525 in-lb/s = 43.75 ft-lb/s = 0.080 hp, less than 0.08% losses;
• There is no differential in a motorcycle, we have said that before;
• The damper should be well into the isolating side of of the forced vibration response curve and be receiving very little energy, particularly if it is rubber, as it will quickly cook if there is any measurable energy going into it. The dampener will have power consumption similar to the swing arm movement energy. I know this topic pretty well, at Ford my title was Technical Specialist for Torque Converter Clutch and Damper;
• U-joints or CV joints in swing arm bikes - Not usually pertinent to airplanes, and I understand there to be only one CV or Hooke joint in most bikes, coincident with the swing arm pivot. It has about the same friction force as at the gear teeth, 960 lb*0.005*radius of the joint*sine(angular swing). At 10 degrees and a 1-3/4" working diameter, that works out to 6.6 ft-lb/second, or about 0.012 hp - insignificant;
• Any downstream clutch is an overload clutch and should not be slipping at all except maybe the I*alpha of the engine is picked up during gear changes, but not on steady pulls;
• Rubber boots have the same thing going on as the the other parts, tiny energy;
• Oil and grease are part of the 0.005 drag coefficient as the parts roll back and forth. If it is a good oil spray lube, this value is usually more like 0.001 - 0.002;
This stuff adds up, but it adds up to less than the losses at the gear mesh and the bearings carrying the loads from the gears.

Yeah, it seems the methods are different. Ok, so how would you calculate the losses from for example a Honda Accord 2.0i from 2007 with 155HP at the crankshaft, one automatic and the other manual, what methods would you use? They both have the same engine, only the drivetrain is different. This was a very common Japanese car in EUrope, and the manufacturer had to comply to EUropean homologation rules.
Nope, you brought nothing to the table but outlandish claims and criticisms of someone explaining how to do it from first principles, and want me to do a big sophisticated analysis?

It is different in the EU. Many EU countries charge taxes by the engine kilowatts, so this is quite strict. Before kilowatts it used to be horsepower until 2010, and then the official measurement became the kilowatt while horsepower was only noted on the side. A kilowatt is (in countries which accept the DIN standard) the exact power necessary to lift 75kg (hence 75 kiloponds) 1 meter high in the time of one second.
Wrong again. 75kg lifted 1 m is 75kg * 9.802 m/s/s * 1 m/s = 735N*m/s which is 735 watts. One horsepower is 746 Watts, and 1000 Watts is a kilowatt.

Ok, so are you saying that you do not verify your drivetrain losses calculation on a dyno after doing all the math? How will you know then that your losses and efficiency are exact?
I started all of this explaining that the transmission builders run efficiency studies as part of verifying and validating the design. I even went so far as to explain that they are run with double ended dynos and both with all hydraulics friction devices running and with pinned friction devices.

Billski

Well-Known Member
I have read the to and fro comments and think the main difference that is not taken into account by dangerzone is tyre loss on a dyno.
Axle mounted dynos prove this.
Hey - that mob are our competitors!
You can't trust them - they eat babies and are all-round knaves and villains.

(But our tests give similar results)

D Hillberg

Well-Known Member
This would depend on the rotor diameter. But if I remeber well from a UH-60 Blackhawk OH, the main rotor was around 250 rpm while the tail was at around 1200 rpm. Neither would be suitable for a homebuilt aircraft.
The 2300 rpm of the S-58 had 150 hp to play with - the tail rotor gear box works fine but has a 2:1 ratio.
The shaft drive for a Goldwing motorcycle works great as a main gearbox for a 60/70 hp single place helicopter.
The Winters racing rear end will do 150 hp all day in a Jet Exec. with a wide choice of ratios.
The UH 60 t/r gear box is no different then any other
Lots of industrial gearboxes with aluminum housings on the market.
Have done multiple helicopters with rotor diameters from 18 ft to 25 feet no issues.

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DangerZone

Well-Known Member
HBA Supporter
No you do not see. Specifically:
• I discussed the bearings on the 3/4" diameter shaft, their loads, the friction forces, the friction torque and then the power used turning them. they were part of the total;
• Swing arms major deflection occurs once when driveshaft torque is increased, and then unwinds when driveshaft torque decreases. So once per throttle cycle or gear change if power is release during the gear change, less when shifted without the clutch. At firing order there should be very low deflections, on the order of thousandths. So if we have 840 in-lb torque and say 0.005" firing order deflection, that is 840 in-lb * 0.005 * 125/s = 525 in-lb/s = 43.75 ft-lb/s = 0.080 hp, less than 0.08% losses;
• There is no differential in a motorcycle, we have said that before;
• The damper should be well into the isolating side of of the forced vibration response curve and be receiving very little energy, particularly if it is rubber, as it will quickly cook if there is any measurable energy going into it. The dampener will have power consumption similar to the swing arm movement energy. I know this topic pretty well, at Ford my title was Technical Specialist for Torque Converter Clutch and Damper;
• U-joints or CV joints in swing arm bikes - Not usually pertinent to airplanes, and I understand there to be only one CV or Hooke joint in most bikes, coincident with the swing arm pivot. It has about the same friction force as at the gear teeth, 960 lb*0.005*radius of the joint*sine(angular swing). At 10 degrees and a 1-3/4" working diameter, that works out to 6.6 ft-lb/second, or about 0.012 hp - insignificant;
• Any downstream clutch is an overload clutch and should not be slipping at all except maybe the I*alpha of the engine is picked up during gear changes, but not on steady pulls;
• Rubber boots have the same thing going on as the the other parts, tiny energy;
• Oil and grease are part of the 0.005 drag coefficient as the parts roll back and forth. If it is a good oil spray lube, this value is usually more like 0.001 - 0.002;
This stuff adds up, but it adds up to less than the losses at the gear mesh and the bearings carrying the loads from the gears.
Actually, you helped a lot, your answers are appreciated. You showed me how you evaluate losses, and how you add them up instead of multiplying them since they are all part of an interconnected drivetrain system. Now I understand why there is so much difference between your calculation and how it is done in the metric world. Btw, most good quality shaft drives have two U joints, because the drives with only one used to break the differential teeth in the past due to swingarm deflection. Yet since neither you nor anybody else here said what you call a motorcycle shaft drive differential, it is hard to say how to call the mechanical piece which often broke in older motorcycle shaft drives. I'll just accept this odd thing that you call it nothing, but many others call it a motorcycle shaft drive differential.

Nope, you brought nothing to the table but outlandish claims and criticisms of someone explaining how to do it from first principles, and want me to do a big sophisticated analysis?
Nope, I am not criticizing, I was merely trying to understand why there is so much difference in how you calculate losses and efficiency. Your answers are greatly appreciated.

Wrong again. 75kg lifted 1 m is 75kg * 9.802 m/s/s * 1 m/s = 735N*m/s which is 735 watts. One horsepower is 746 Watts, and 1000 Watts is a kilowatt.
Ok, metric horsepower is wrong, imperial horsepower is right.
Horsepower - Wikipedia
This is exactly why it is good that power is expressed officially in kilowatts after 2010 in the EU instead of horsepower.

I started all of this explaining that the transmission builders run efficiency studies as part of verifying and validating the design. I even went so far as to explain that they are run with double ended dynos and both with all hydraulics friction devices running and with pinned friction devices.

Billski
Fair enough, same here. You probably also mark the tires to account for tire slip just like nuts and bolts are marked in aviation. We often do a reality check after all that and use physics to check the power empirically in real conditions on the track. It is often like this for high power bikes:

For example, these crankshaft+200HP bikes often show only 188.4HP in real life conditions like in this video, and they are really well tuned for maximum performance and least power losses. The same can be done (and was done) for shaft driven motorcycles in the whole power range, or any vehicle. So far, physics always show the real numbers.

DangerZone

Well-Known Member
HBA Supporter
Btw, on 2 wheel motorcycles with shaft drive, there is NO differential.
Ok, I could accept this is a language thing, but then again:

No differentials? Ok, so what would you call the rear shaft drive differentials that are sold on ebay when a shaft drive differential breaks due to hard abuse? It looks like this:

1997-2001 BMW K1200LT K1200 NON-ABS Rear Wheel Drive Axle Differential Gear Case | eBay

1975 MOTO GUZZI 850 T 850T REAR DIFFERENTIAL | eBay

1991 SUZUKI KATANA GSX1100 G GSX 1100g FINAL DRIVE REAR DIFFERENTIAL 89 90 91 | eBay
Nobody answered? Nobody here knows what you call the motorcycle shaft drive differential? If it breaks and you need to buy one, do you ask for "the thing that is not a differential, but looks and performs as one in some motorcycle shaft drives"?

Kiwi303

Well-Known Member
I owned a car that had the rear disk brakes mounted on the inner end of the half-shafts, right beside the differential.
Old Jaguars were a B***h to maintain... the models with that setup happened right in the Commie Strikes period of British unionism. Add infamously unreliable Lucus electrics made by blind chimps wearing boxing gloves and a recipe for disaster!

TiPi

Well-Known Member
Log Member
Ok, I could accept this is a language thing, but then again:

Nobody answered? Nobody here knows what you call the motorcycle shaft drive differential? If it breaks and you need to buy one, do you ask for "the thing that is not a differential, but looks and performs as one in some motorcycle shaft drives"?
you can call it a final drive, an angle drive or rear axle drive

Well-Known Member
Ok, I could accept this is a language thing, but then again:

Nobody answered? Nobody here knows what you call the motorcycle shaft drive differential? If it breaks and you need to buy one, do you ask for "the thing that is not a differential, but looks and performs as one in some motorcycle shaft drives"?
But it doesn't do anything like a differential!

A differential allows a single input shaft to drive two output shafts with different speeds.

DangerZone

Well-Known Member
HBA Supporter
you can call it a final drive, an angle drive or rear axle drive
Are you joking?

But it doesn't do anything like a differential!

A differential allows a single input shaft to drive two output shafts with different speeds.
Are you kidding me?

So, if you want to buy this part you go to a store and ask for "a thing that doesn't do anything like a differential"?

Guys, it's a simple question and it would be nice to know what all of you here on this forum call it. The first seller is from Virginia, the second from Florida, the third is from Texas. So it is not just us Europeans who call it a motorcycle rear shaft drive differential.

Geraldc

Well-Known Member
It seems that a lot of motorcycle dealers call the gear housing that contains the crown wheel and pinion a differential even though there are no spider gears and only one output.

DangerZone

Well-Known Member
HBA Supporter
Just how much power do you thing swingarm deflection could absorb? What differential would you have in a bike, and in any case how does a differential lose power if both shafts are being turned at the same rate so that the spider gears are not turning? How does a clutch lose power when it isn't slipping? How do U-joints lose power if their shafts are lined up? How do rubber boots absorb power when they're not even turning?

I think you're well out in left field here and don't understand that HP is torque times RPM, and to lose much HP through efficiency losses you have to have some serious heat-generating friction somewhere. Swingarm deflection, for instance, isn't going to create enough frictional heat to absorb even half a watt. U-joints lose nearly nothing, even if the shafts are angled at 10° . Their needle bearings are pretty efficient. Rubber boots absorbing HP? Come on. On and on it goes. Like has been said multiple times here, you'd be coking the oil in that thing, or melting it down, if it was it was losing 18 HP. That's serious kilowattage there. Physics and math, every time.
A very little bit. However, there are a lot of such small losses of each component in a motorcycle shaft drive which influence one another in series.

Heat dissipates. Slipping can have more or less fricition. However, physics and math work every time and it seems we both agree on that. I posted here two methods to determine power losses in a motorcycle from crankshaft to the rear tire. One is a calcualtion which is taught at school, the other is a video of Matt Hudson who is a great bike enthusiast.These methods are not mine, people use them all the time. So instead of arguing with me that this is against math and physics, why would you not try to calculate it yourself?

I honestly don't have all the answers to where does the heat go, how come there is so much loss, what is the meaning of life, etc. I'm a simple guy, I measure things, calculate the results, verify the results with a double reality check and if the math works - ok. Most of you argue this is serious wattage. Sure, it is. However, this will not change the fact that these losses are real. In fact, I would LOVE to have a motorcycle transmission drive with less than 1% losses, I would really LOVE it. But I have to live in the real world and accept reality with all joys and losses.