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Mechanical Efficiency
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#13
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Mechanical Efficiency
On Fri, 21 Apr 2017 08:08:06 -0700 (PDT), wrote:
I had started this because in a recumbent the chain run is so long that it can't possibly be very efficient. I beg to differ. Long chains are more efficient than short chains. The loss of efficiency in a chain drive is mostly from the friction losses when the chain link pin is rotated inside the sleeve. Only the upper part of the chain loop is under tension. Only the links that are rotating from the straight line chain, to the crank and the freewheel have pins that are rotating and therefore add friction. In other words, for large diameter gears, only two links on the entire chain have rotating pins. For smaller diameters and tighter bends, perhaps 2 more pins. In addition, the tension on the chain is distributed only over the upper part of the chain loop. For example, if you a pulling with 100 lbs of tension on the upper part of the chain loop, and there are 10 links (and 20 pins) in the upper part of the chain loop, then the pressure per pin is: 100 lbs / 20 pins = 5 lbs/pin However, if you use a much longer chain, as in a recumbent, with 50 links (100 pins) between the two gears, the tension per pin is: 100 lbs / 100 pins = 1 lbs/pin Since frictional losses increase with surface pressure, the longer chain would have LESS friction than the shorter chain because there is less pressure on the pins and sleeves. Of course, a longer chain would weigh more, but that's another calculation and is not directly involved in the efficiency calculation. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
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Mechanical Efficiency
On Friday, April 21, 2017 at 8:33:54 AM UTC-7, Frank Krygowski wrote:
On 4/21/2017 11:08 AM, wrote: I had started this because in a recumbent the chain run is so long that it can't possibly be very efficient. I don't know of any reason that a long chain should be much less efficient than a short one. Chain inefficiencies (small as they are) come from the friction as the links bend around sprockets, and friction as rollers roll on and off sprocket teeth. Some of the bending friction is between side plates exacerbated by lateral angles, and long chain runs tend to have smaller lateral angles. They may be more efficient than short chains. -- - Frank Krygowski Long chains have several guide pulleys along their length and each one of those pulleys absorb about a watt each. |
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Mechanical Efficiency
On 4/21/2017 10:33 AM, Frank Krygowski wrote:
On 4/21/2017 11:08 AM, wrote: I had started this because in a recumbent the chain run is so long that it can't possibly be very efficient. I don't know of any reason that a long chain should be much less efficient than a short one. Chain inefficiencies (small as they are) come from the friction as the links bend around sprockets, and friction as rollers roll on and off sprocket teeth. Some of the bending friction is between side plates exacerbated by lateral angles, and long chain runs tend to have smaller lateral angles. They may be more efficient than short chains. Welp,,,, I would presume that a longer chain would tend to require more idlers to route a circuitous path than a short one would. Using plain bicycle chain still tends to win out most of the time tho. It's just easier and cheaper to use than anything else. Home-builders can make sprockets any size you want with just a drill press and any 2-D CAD program. Any of that other stuff is way more complicated to try to fabricate. I have pondered the same problem too however: with a conventional bicycle drive train, even with triple chainrings you run into the problem of not having enough width in the drive ratios... You want to be able to crawl in a granny gear when you need to, but you still want to be able to pedal at a comfortable cadence at 35 - 40+ MPH. ,,,,,,, I know how most do it: they just use a normal 2x9+ setup with the widest ranges they can get, and they sacrifice the low end for a bit higher end. Some bikes also use compound gearing, with a second rear hub + derailler after the first, but that gets messy fast. And none of these is still really convenient to use, in that they do not place the ratios in a mathematical order. Ideally you would want a couple of button-levers marked [shift up] and [shift down], and it would go through all the ratios in order, from one end to the other. With an automatic car transmission for example, it shifts 1-2-3-4-5 automatically when accellerating. You don't need to TELL it what gear to use..... I've seen this same question asked about Di2 setups: why is there still two levers? Since a microprocessor could easily figure out the next-highest gear or next-lowest gear, and switch the front & rear deraillers to reach it. You should just need ONE lever, or two buttons.... But this only seems to be stubborn tradition; riders are used to mentally managing the front and rear deraillers separately. So that was the way Shimano decided to control it. |
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Mechanical Efficiency
On Friday, April 21, 2017 at 10:12:39 AM UTC-7, Jeff Liebermann wrote:
On Fri, 21 Apr 2017 08:08:06 -0700 (PDT), wrote: I had started this because in a recumbent the chain run is so long that it can't possibly be very efficient. I beg to differ. Long chains are more efficient than short chains. The loss of efficiency in a chain drive is mostly from the friction losses when the chain link pin is rotated inside the sleeve. Only the upper part of the chain loop is under tension. Only the links that are rotating from the straight line chain, to the crank and the freewheel have pins that are rotating and therefore add friction. In other words, for large diameter gears, only two links on the entire chain have rotating pins. For smaller diameters and tighter bends, perhaps 2 more pins. In addition, the tension on the chain is distributed only over the upper part of the chain loop. For example, if you a pulling with 100 lbs of tension on the upper part of the chain loop, and there are 10 links (and 20 pins) in the upper part of the chain loop, then the pressure per pin is: 100 lbs / 20 pins = 5 lbs/pin However, if you use a much longer chain, as in a recumbent, with 50 links (100 pins) between the two gears, the tension per pin is: 100 lbs / 100 pins = 1 lbs/pin Since frictional losses increase with surface pressure, the longer chain would have LESS friction than the shorter chain because there is less pressure on the pins and sleeves. Of course, a longer chain would weigh more, but that's another calculation and is not directly involved in the efficiency calculation. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 Huh? These long chains do not have straight runs and in fact near the back have a rather shorter run to the cogs which always puts them in more of a sharper angle to the cogset. In other words - you are always in a more "cross-chained" angle. |
#17
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Mechanical Efficiency
On Friday, April 21, 2017 at 10:12:39 AM UTC-7, Jeff Liebermann wrote:
On Fri, 21 Apr 2017 08:08:06 -0700 (PDT), wrote: I had started this because in a recumbent the chain run is so long that it can't possibly be very efficient. I beg to differ. Long chains are more efficient than short chains. The loss of efficiency in a chain drive is mostly from the friction losses when the chain link pin is rotated inside the sleeve. Only the upper part of the chain loop is under tension. Only the links that are rotating from the straight line chain, to the crank and the freewheel have pins that are rotating and therefore add friction. In other words, for large diameter gears, only two links on the entire chain have rotating pins. For smaller diameters and tighter bends, perhaps 2 more pins. In addition, the tension on the chain is distributed only over the upper part of the chain loop. For example, if you a pulling with 100 lbs of tension on the upper part of the chain loop, and there are 10 links (and 20 pins) in the upper part of the chain loop, then the pressure per pin is: 100 lbs / 20 pins = 5 lbs/pin However, if you use a much longer chain, as in a recumbent, with 50 links (100 pins) between the two gears, the tension per pin is: 100 lbs / 100 pins = 1 lbs/pin Since frictional losses increase with surface pressure, the longer chain would have LESS friction than the shorter chain because there is less pressure on the pins and sleeves. Of course, a longer chain would weigh more, but that's another calculation and is not directly involved in the efficiency calculation. So you are saying that losses from routing the lower run of the chain through tensioners and idlers are negligible? |
#18
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Mechanical Efficiency
On Fri, 21 Apr 2017 20:03:26 -0400, Frank Krygowski
wrote: On 4/21/2017 1:12 PM, Jeff Liebermann wrote: On Fri, 21 Apr 2017 08:08:06 -0700 (PDT), wrote: I had started this because in a recumbent the chain run is so long that it can't possibly be very efficient. I beg to differ. Long chains are more efficient than short chains. The loss of efficiency in a chain drive is mostly from the friction losses when the chain link pin is rotated inside the sleeve. Only the upper part of the chain loop is under tension. Only the links that are rotating from the straight line chain, to the crank and the freewheel have pins that are rotating and therefore add friction. In other words, for large diameter gears, only two links on the entire chain have rotating pins. For smaller diameters and tighter bends, perhaps 2 more pins. In addition, the tension on the chain is distributed only over the upper part of the chain loop. For example, if you a pulling with 100 lbs of tension on the upper part of the chain loop, and there are 10 links (and 20 pins) in the upper part of the chain loop, then the pressure per pin is: 100 lbs / 20 pins = 5 lbs/pin However, if you use a much longer chain, as in a recumbent, with 50 links (100 pins) between the two gears, the tension per pin is: 100 lbs / 100 pins = 1 lbs/pin Not true, Jeff. The tension in the chain is a constant The load on each pin in the free upper span is exactly the same, and it doesn't change if the chain is longer, i.e. has more pins. A normal chain with a 100 pound load has 100 pounds on each pin, no matter how long. Argh. We went through this exercise a few years ago in this newsgroup when I allegedly made the same mistake. Am I wrong again? (I just hate it when that happens). The way I look at it is that if I replace each link in the upper part of the chain loop with a spring scale, methinks the deflection of each spring (a measure of the force) would be the pulling load divided by number of spring scales. Or, if we break the chain and put one spring scale between two adjacent links, the measured force will be equal to the applied load. However, if we break the chain in two places, methinks the measured force will be half the applied load. If the force were equal to the applied load on each link, I would expect the two spring scales to also indicate a force equal to the applied load, which I don't believe is the case. Now if you had two chains side by side sharing the same 100 pound load, those pins would each have 50 pounds on them. Yep, here we agree. To use an electrical analogy, you're confusing a series situation with a parallel situation. The electrical analogy would be each link in the chain is a resistor. The applied load is the voltage. If the resistors are all equal, the voltage across each resistor would be equal to the applied voltage divided by the number of resistors. I'll admit that I might be wrong about all this, but at this moment, it seems correct to me. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
#19
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Mechanical Efficiency
On Fri, 21 Apr 2017 10:40:30 -0700 (PDT), Doug Landau
wrote: So you are saying that losses from routing the lower run of the chain through tensioners and idlers are negligible? Worse. I was ignoring the losses from tensioners, idlers, grease, pin rotation caused by a slight chain droop, inertial loading from a longer and thus heavier chain, chain acceleration lag, etc. I assumed that the original question was about a simple power transmission system, not the complex mess that such systems inevitably evolve into. Something more like this test fixture and a practical bicycle: http://cdn.mos.bikeradar.imdserve.com/images/news/2012/11/06/1352163122826-1476emv18vmdi-630-80.jpg The additional losses can be tested separately and included later. You're correct that tensioners, idlers, etc are important. For a chain drive that is about 95% efficient, small frictional losses might reduce the 95% efficiency rather significantly. However, I don't believe we were trying to nail down a single number for a chain drive efficiency. As noted in the "Bicycling Science" books, the efficiency will vary somewhat depending on speed, conditions, maintenance, gearing, etc. What I believe was of interest was comparing the efficiencies of a drive chain, hydraulic drive, and an electric generator-motor drive. From my readings, these will show drastically different efficiencies, where the effects of tensioners, idlers, etc will not make much difference in the comparison. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
#20
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Mechanical Efficiency
On Fri, 21 Apr 2017 10:18:17 -0700 (PDT), wrote:
Huh? These long chains do not have straight runs and in fact near the back have a rather shorter run to the cogs which always puts them in more of a sharper angle to the cogset. In other words - you are always in a more "cross-chained" angle. There are chain designs which help in such arrangements. This one seems popular: http://kmcchain.us/chain/x9-93/ http://t-cycle.com/chain-in-bulk-c-116/bulk-kmc-x993-chain-by-foot-p-35.html However, if you're looking for the source of chain drag and friction, just look at what wears on an old chain. It's not the inner or outer plates, which is what allegedly wears with "bent" chain line. It's the pin and sleeve that show most of the wear and presumably is the source of most of the friction. I say presumably because there are differences in surface hardness, lubrication, cleanliness, etc which affect wear. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
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