Mono Posted August 10, 2017 Share Posted August 10, 2017 EDIT: you may want to check out these links http://forum.electricunicycle.org/topic/7549-current-demand-versus-battery-voltage/?page=3#comment-106384http://forum.electricunicycle.org/topic/7855-anatomy-of-an-overlean/http://www.ebikes.ca/tools/simulator.html as they provide similar and IHMO even more informative graphs. I collected a few graphs based on a relatively simple common motor model and calculations of drag and rolling resistance from these sources: https://evmc2.wordpress.com/2014/07/21/electric-motor-power-really-simple-and-hp-ratings/http://lancet.mit.edu/motors/motors3.htmlhttp://www.me.mtu.edu/~wjendres/ProductRealization1Course/DC_Motor_Calculations.pdfhttps://www.gribble.org/cycling/power_v_speed.htmlhttp://www.engineeringtoolbox.com/drag-coefficient-d_627.htmlhttps://en.wikipedia.org/wiki/Body_surface_area All errors are mine and bound to be corrected after a good nights sleep. The parameters I have set are: overall weight: 100kg rolling coeffient: 0.01 drag coefficient: 1.0 frontal area: 0.8m^2 max motor speed: 50-60km/h motor power: 500-1200W For a basic orientation, these are power versus speed graphs for an 800W motor: According to the motor model, the maximally available torque decreases linearly with increasing speed and becomes zero at max speed (as shown in the below figure, and like the shown estimated input power graph). Max speed is 60km/h here. From this, the shown mechanical power computes straight forward to be proportional to speed times torque. The mechanical power graph (orange line) has therefore a parabolic shape, where the maximal power output of the motor occurs at 1/2 of its limit speed, here at 30km/h. At 1/2 of the max speed, the electric input power is 1600W, the output power is 800W, the efficiency is 50%. The necessary electric input power to produce maximal torque grows to 4 times the nominal motor power at zero speed, here 3200W. The motor efficiency (all at max load) becomes the smaller the smaller the speed. Therefore, battery limitation may limit the shown electrical input and the mechanical output power to smaller values than displayed. (Not easily applicable to BLCD motors). The practical (very unsafe) limit speed is around 39km/h, where the power needed to keep up the speed (red line) crosses the motor power. Above 20km/h, the power needed to keep up speed is dominated by drag. More interesting is the forward thrust (force) of the same 800W motor due to its torque. (The torque is determined by its max speed and power specs): The nominal torque/forward thrust is just a straight line from its maximal value down to zero at max speed. The used unit for the y-axis is the downward force of one kg (i.e. 9.81 N). This is a rather tangible unit, as we have a feeling for the downforce, that is the weight, of a kg. At 30km/h, almost half of the torque is needed to keep up the speed. An equivalent of 5kg additional "freely available" surplus thrust remains to accelerate or balance the wheel, to get over bumps or out of potholes etc. A little scary, but OK-ish. Even more interesting (in light of http://forum.electricunicycle.org/topic/7654-story-of-me-learning-riding-and-falling-off-an-inmotion-v8) is investigating the "free" remaining surplus forward thrust (force) to accelerate and balance the wheel as a function of speed for different slopes. The surplus thrust is shown for three different motor specs and three different slopes. The vertical distance between the graphs of the same color is pretty much proportional to overall weight (here 100kg). This shows that riding into a slope with high speed is crazy dangerous. Driving at 20km/h into a 10% = 5.7º incline, an 800W 55km/h motor drops from 11kg to 1.1kg surplus-thrust to just about still balance the rider. (A 800W 40km/h motor has 2.2kg left, which is only little better). For a slope of 0% the graphs are almost independent of weight (only rolling resistance depends on weight). At 30km/h, any rider of a 500/800/1200W wheel has <2/5/10kg to play with, whether they weigh 60kg or 120kg. The only difference is that 2/5/10kg feels less to play with for a 120kg rider. The critical lean angle depends of course on the weight, as much at 30km/h as it does at 0km/h. Finally the same graphs assuming a limiter restricts the current to three times the nominal motor power, e.g. to 30A = 3 * 800W / 80V. (Limiting to four times the nominal motor power doesn't change the above graphs.) I can do these graphs easily for any parameter configuration, if you ask I may give it a shot... Link to comment Share on other sites More sharing options...
LanghamP Posted August 12, 2017 Share Posted August 12, 2017 A person who uses the possessive "its" properly instead of "it's" is worthy of the greatest respect. Can you estimate wattage for a 100kg rider going up slopes at between 8-13 mph? I get in my opinion ubsurdly low rates like between 400-600 watts, and often much less. Link to comment Share on other sites More sharing options...
Mono Posted August 12, 2017 Author Share Posted August 12, 2017 shows mechanical power needed for slopes up to 15% for 120kg wheel+rider, where 8-13mph is pretty much 13-21km/h. Now, I am not sure how to estimate efficiency under these operating conditions. Using 1 - torque/torque_max as efficiency coefficient (EDIT: which is likely to be very inaccurate), we get for a 1200W motor (best case) At 5% slope, we see 400-700W for the above mentioned speed range, but this becomes quickly much more with increasing slope. Link to comment Share on other sites More sharing options...
meepmeepmayer Posted August 12, 2017 Share Posted August 12, 2017 This is too much for me today, but... Will have a closer look soon... Link to comment Share on other sites More sharing options...
palachzzz Posted August 13, 2017 Share Posted August 13, 2017 Is there anything to back but theory? On your graphs, the maximum power of V8 is taken for 800 watts, look at this article https://airwheel.ru/test-monokoles-na-dinostende/, but it is in Russian, you can use google for translate https://translate.google.ru/translate?sl=ru&tl=en&js=y&prev=_t&hl=ru&ie=UTF-8&u=https%3A%2F%2Fairwheel.ru%2Ftest-monokoles-na-dinostende%2F&edit-text=. Inmotion V8 on the dyno shows 2,2 kW of mechanical power on the default battery, and 2.7 kW of mechanical power on a high-performance battery. In addition, in half of the articles you refer to the characteristics of brushed DC motor, but we have a sensored brushless motor. It's completely different motors with completely different characteristics. Link to comment Share on other sites More sharing options...
Mono Posted August 13, 2017 Author Share Posted August 13, 2017 20 hours ago, palachzzz said: Is there anything to back but theory? As Feynman once famously said, "If it [the law] disagrees with experiment, it's wrong. And that simple statement is the key to science. It doesn't make a difference how beautiful your guess is. It doesn't make a difference how smart you are, who made the guess, or what his name is; if it disagrees with experiment: wrong. That's all there is to it." The only "theory" specific to electric motors that is used is the linearly decreasing dependency between torque and speed. The maximally available torque is maximal at zero speed and goes linearly down to zero at maximal speed (that's the reason why this is the maximal speed, there is zero torque available to increase speed). This graphs is shown in the second figure. Everything else is derived from this torque-vs-speed relation and high-school physics unrelated to EUCs. If you know a better model for the torque vs speed dependency of electric motors bring it on, I am always happy to learn. 20 hours ago, palachzzz said: Inmotion V8 on the dyno shows 2,2 kW of mechanical power on the default battery, and 2.7 kW of mechanical power on a high-performance battery. Interesting point that an 800W motor can deliver 2.7kW of mechanical power. Maybe it can for a short period of time, while 800W is the power it can sustain permanently. This may indeed render the above estimates for torque reserves for balancing, bumps or potholes too pessimistic. 20 hours ago, palachzzz said: In addition, in half of the articles you refer to the characteristics of brushed DC motor, but we have a sensored brushless motor. It's completely different motors with completely different characteristics How does the basic physics of an electric motor depend on whether it uses brushes? Link to comment Share on other sites More sharing options...
Slaughthammer Posted August 13, 2017 Share Posted August 13, 2017 On 12.8.2017 at 6:39 PM, Mono said: Using 1 - torque/torque_max as efficiency coefficient, we get for a 1200W motor (best case) Can you please explain how this formula is derived? Link to comment Share on other sites More sharing options...
Mono Posted August 13, 2017 Author Share Posted August 13, 2017 12 minutes ago, Slaughthammer said: Can you please explain how this formula is derived? It is a guess. For maximal load and up to 90% speed, the efficiency coefficient 1 - torque/torque_max represents the data presented in http://www.me.mtu.edu/~wjendres/ProductRealization1Course/DC_Motor_Calculations.pdf and elsewhere very well. As I wrote before I gave the formula, I am not sure how to estimate efficiency under other operating conditions. The formula is obviously wrong at zero speed, where efficiency is always zero. Link to comment Share on other sites More sharing options...
Slaughthammer Posted August 13, 2017 Share Posted August 13, 2017 Well... might want to take a look here: (http://www.solar-wind.co.uk/solar-water-pumps-borehole-well-pumping-irrigation.html) Link to comment Share on other sites More sharing options...
Mono Posted August 13, 2017 Author Share Posted August 13, 2017 17 hours ago, Slaughthammer said: Well... might want to take a look here: (http://www.solar-wind.co.uk/solar-water-pumps-borehole-well-pumping-irrigation.html) Thanks for the link! The difference is, AFAICS, that they can choose the speed, so it seems reasonable to assume they chose the speed with optimal efficiency (for high loads about 80-90% of max speed). For EUCs we want to know the efficiency for any given speed. Still helpful! My reasoning was that at low speed and high load, the efficiency is very low. Now, what about low load? At low constant speed (ie. low load) we tend to see the largest ranges per Wh, hence rather high than low efficiency. Link to comment Share on other sites More sharing options...
palachzzz Posted August 13, 2017 Share Posted August 13, 2017 4 hours ago, Mono said: Interesting point that an 800W motor can deliver 2.7kW of mechanical power. Maybe it can for a short period of time, while 800W is the power it can sustain permanently. This may indeed render the above estimates for torque reserves for balancing, bumps or potholes too pessimistic. Why do you think that in V8 800 watts? Simply because it is described as the nominal power? But why don't think that this nominal power means the possibilities of a battery? It can not provide high currents for long time - otherwise it will be overheating, I think the power 800 watts (~80 v at 10 amps) is taken as the nominal for the whole EUC, but the weak point of it is battery, not motor. Link to comment Share on other sites More sharing options...
Mono Posted August 13, 2017 Author Share Posted August 13, 2017 14 hours ago, palachzzz said: Why do you think that in V8 800 watts? Simply because it is described as the nominal power? But why don't think that this nominal power means the possibilities of a battery? Because 800W is given as motor specs, not as battery specs. I do understand that battery may also be a limiting factor, of course. Quote It can not provide high currents for long time - otherwise it will be overheating, I think the power 800 watts (~80 v at 10 amps) is taken as the nominal for the whole EUC, but the weak point of it is battery, not motor. The V8 has a 460Wh battery. Discharging at 2C^1, which is to my understanding entirely within the standard range of operation for Li-ion batteries, yields 920W. It would be quite funny if overheat would determine the power specs of EUCs, because that would mean their power specs were determined by the size of their heat sinks. I don't think so. ^1 http://batteryuniversity.com/learn/article/discharge_characteristics_li Link to comment Share on other sites More sharing options...
palachzzz Posted August 13, 2017 Share Posted August 13, 2017 13 minutes ago, Mono said: Because it is given as motor specs, Please, show me the place where it is written that this is the power of the motor. 13 minutes ago, Mono said: The V8 has a 460Wh battery. Discharging at 2C^1, which is to my understanding entirely within the standard range of operation for Li-ion batteries, yields 920W. And again references to mythical and theoretical figures. There is the test of LG MH1 that used in V8: https://www.e-cigarette-forum.com/forum/threads/lg-mh1-10a-3200mah-18650-bench-test-results-safe-at-10a-but-suffers-damage.686757/ Link to comment Share on other sites More sharing options...
Mono Posted August 13, 2017 Author Share Posted August 13, 2017 4 minutes ago, palachzzz said: Please, show me the place where it is written that this is the power of the motor. Are you serious? Link to comment Share on other sites More sharing options...
Mono Posted August 13, 2017 Author Share Posted August 13, 2017 9 hours ago, palachzzz said: 10 hours ago, Mono said: The V8 has a 460Wh battery. Discharging at 2C^1, which is to my understanding entirely within the standard range of operation for Li-ion batteries, yields 920W. And again references to mythical and theoretical figures. Sure, maybe you also better tell that Jason, he computes 1440W: sources: https://www.ewheels.com/electric-unicycle-ultimate_ewheel_comparison_ips_airwheel_ninebot_king_song_gotway/https://www.ewheels.com/choosing-an-electric-unicycle-with-the-right-battery-pack-for-you/ It may be that 80V * 10A = 800W, but the V8 has two packs in parallel, hence we need to compute 80V * 10A * 2 = 1600W, or (as recommended in the linked article) 80V * 7A * 2 = 1120W. Or more realistically as Jason 72V * 10A * 2 = 1440W or 72V * 7A * 2 = 1008W. Of course these are all only theoretical figures and they are all very mystical and mysteriously all larger than the 920W I computed to be within the standard range of operation. Link to comment Share on other sites More sharing options...
Chriull Posted August 14, 2017 Share Posted August 14, 2017 14 hours ago, palachzzz said: Is there anything to back but theory? On your graphs, the maximum power of V8 is taken for 800 watts, look at this article https://airwheel.ru/test-monokoles-na-dinostende/, but it is in Russian, you can use google for translate https://translate.google.ru/translate?sl=ru&tl=en&js=y&prev=_t&hl=ru&ie=UTF-8&u=https%3A%2F%2Fairwheel.ru%2Ftest-monokoles-na-dinostende%2F&edit-text=. Inmotion V8 on the dyno shows 2,2 kW of mechanical power on the default battery, and 2.7 kW of mechanical power on a high-performance battery. The 2.2/2.7 kW numbers are the absolute maximal output powers the V8 can produce with full batteries and cold motor. Also the 2.7kW are already a bit "over the maximum" - @EcoDriftblew the fuse once measuring the modified V8. He measured the same numbers for the KS16C - which is also announced to have a nominal power of 800W. The measured 2.2kW max output power seem quite feasable with full batteries and cold motor: http://forum.electricunicycle.org/topic/7549-current-demand-versus-battery-voltage/?do=findComment&comment=106424 - just the on some sites advertised 3kW peak power are way out of reach... On the other side, in a real world situation (somewhat used up batteries, warm motor - http://forum.electricunicycle.org/topic/7855-anatomy-of-an-overlean/) the limit shifted down to somewhere around ~1.5kW (also i don't know how valid and accurate the numbers reported from the wheel are...) So @Mono's 800W can be reached anytime with an V8, maybe also still something around 1kW - with not too empty batteries something around this 1.5kW and with full batteries one could reach the 2.2kW... 14 hours ago, palachzzz said: In addition, in half of the articles you refer to the characteristics of brushed DC motor, but we have a sensored brushless motor. It's completely different motors with completely different characteristics. Brushed DC and BLDC motors function on the same principles - for our discussions here this simplification is totaly valid. Brushes or not, number of coils, etc has no influence on this principles. Link to comment Share on other sites More sharing options...
Chriull Posted August 14, 2017 Share Posted August 14, 2017 On 11.8.2017 at 1:28 AM, Mono said: ... At 1/2 of the max speed, the electric input power is 1600W, the output power is 800W, the efficiency is 50%. The necessary electric input power to produce maximal torque grows to 4 times the nominal motor power at zero speed, here 3200W. The motor efficiency (all at max load) becomes the smaller the smaller the speed. Therefore, battery limitation may limit the shown electrical input and the mechanical output power to smaller values than displayed. ... As the controller together with the motor coil works as DC/DC step down converter (for the BLDC as DC/AC step down converter) the efficiency is much higher - one does not have to burn half of the battery power to reduce the voltage to (almost) 1/2... The main losses at the motor side are just the power "burned" at the ohmic coil resistance, magnetic losses and friction. Then there is the efficiency of the step down converter and the losses in the battery (internal resistance). Link to comment Share on other sites More sharing options...
palachzzz Posted August 14, 2017 Share Posted August 14, 2017 1 hour ago, Chriull said: As the controller together with the motor coil works as DC/DC step down converter (for the BLDC as DC/AC step down converter) the efficiency is much higher - one does not have to burn half of the battery power to reduce the voltage to (almost) 1/2... The main losses at the motor side are just the power "burned" at the ohmic coil resistance, magnetic losses and friction. Then there is the efficiency of the step down converter and the losses in the battery (internal resistance). So, If we consider the efficiency of the system as 80% (as confirmed by the dyno), and take the voltage / current logs in the system, do you agree that we can to know the approximately power of V8 in real world? 1 hour ago, Chriull said: @EcoDriftblew the fuse once measuring the modified V8. Yes, 2,7kW it is about 3,4 kW of electric power, it is about 45A - too much for 30A fuse in V8. But again, the motor not a weak point of the system. Link to comment Share on other sites More sharing options...
Chriull Posted August 14, 2017 Share Posted August 14, 2017 14 minutes ago, palachzzz said: So, If we consider the efficiency of the system as 80% (as confirmed by the dyno), and take the voltage / current logs in the system, do you agree that we can to know the approximately power of V8 in real world? Efficiency is a function of speed and load. But beside of this we have with the 2.2kW a "certified" (approximate) value of the maximal power the V8 can ouput with full batteries. 14 minutes ago, palachzzz said: Yes, 2,7kW it is about 3,4 kW of electric power, it is about 45A - too much for 30A fuse in V8. But again, the motor not a weak point of the system. The motor is a main part of the "weak points" in the system. The combination Battery (voltage, internal resistance) and motor (kv - Volts/revolutions per second, coil resistance) determine the limits - one can have an overlean witht the V8 with full batteries, where it would need (much) less than this 2.2kW. Battery current is for overleans (reaching the torque limit) at higher speed normaly not the limiting factor. Link to comment Share on other sites More sharing options...
Mono Posted August 14, 2017 Author Share Posted August 14, 2017 2 hours ago, Chriull said: As the controller together with the motor coil works as DC/DC step down converter (for the BLDC as DC/AC step down converter) the efficiency is much higher - one does not have to burn half of the battery power to reduce the voltage to (almost) 1/2... The main losses at the motor side are just the power "burned" at the ohmic coil resistance, magnetic losses and friction. Then there is the efficiency of the step down converter and the losses in the battery (internal resistance). I see, you are saying that Table 1 and the efficiency Graph 1 here http://www.me.mtu.edu/~wjendres/ProductRealization1Course/DC_Motor_Calculations.pdf do not apply in brushless motors. Are you aware of reference figures for BLDC you can point us to? Link to comment Share on other sites More sharing options...
palachzzz Posted August 14, 2017 Share Posted August 14, 2017 9 minutes ago, Chriull said: Battery current is for overleans (reaching the torque limit) at higher speed normaly not the limiting factor. And yes, and not. When the current is increased to the limit values for the battery, the voltage in the system dramatically decreases, thus the electric power is also reduced. For example, at a current 40A on default battery the voltage drop can exceed 1V per element (see discharge graphs), i.e. down to 60 volts for a battery, this can cause even a battery shutdown by BMS, or EUC controller. The EUC is very dynamic system. Take a look at the log for my today 17,5 km commute, peak electric power is 1852.88 watts at speed ~22 km/h, and about 600-800 watts at cruise speed (28-29 km/h) on a flat surface 2017_08_14_09_04_25_cutted.csv Link to comment Share on other sites More sharing options...
Mono Posted August 14, 2017 Author Share Posted August 14, 2017 48 minutes ago, palachzzz said: When the current is increased to the limit values for the battery, the voltage in the system dramatically decreases, thus the electric power is also reduced. Right. Only that high speed overleans do not occur at limit currents of the battery. The higher the speed, the more resistance the battery sees from the back EMF of the motor, hence the smaller is the current we can possibly see. Generally speaking, low speed overleans are those caused by battery weakness or heat problems and high speed overleans are caused by the physical limitations of electric motors. Link to comment Share on other sites More sharing options...
Mono Posted August 14, 2017 Author Share Posted August 14, 2017 I hadn't seen this If I had, I wouldn't have started this thread in the first place @Chriull was faster than me and did also much nicer graphs, IMHO. Link to comment Share on other sites More sharing options...
Chriull Posted August 14, 2017 Share Posted August 14, 2017 17 minutes ago, Mono said: I hadn't seen this If I had, I wouldn't have started this thread in the first place Fits good together - but still a little bit another "style" of graph. But i still have some probs - the current need for my "acceleration" lines to not correspond to reality (http://forum.electricunicycle.org/topic/7855-anatomy-of-an-overlean/?do=findComment&comment=107766) and i did not find my fault till now... (also had not the chance till now to invest some real time into that...;( ) Link to comment Share on other sites More sharing options...
Chriull Posted August 14, 2017 Share Posted August 14, 2017 51 minutes ago, palachzzz said: And yes, and not. When the current is increased to the limit values for the battery, the voltage in the system dramatically decreases, thus the electric power is also reduced. For example, at a current 40A on default battery the voltage drop can exceed 1V per element (see discharge graphs), i.e. down to 60 volts for a battery, this can cause even a battery shutdown by BMS, or EUC controller. That's why i don't really like 2p battery systems The 4p as used with the KS16 is quite the lower limit for my "comfort level"... ... and at low battery levels one has to drive cautious anyway ... 51 minutes ago, palachzzz said: The EUC is very dynamic system. Take a look at the log for my today 17,5 km commute, peak electric power is 1852.88 watts at speed ~22 km/h, and about 600-800 watts at cruise speed (28-29 km/h) on a flat surface 2017_08_14_09_04_25_cutted.csv The power values are unfortionately not valid - they come from multiplying motor current times battery voltage with makes no (real) sense. For more details on this topic see http://forum.electricunicycle.org/topic/7549-current-demand-versus-battery-voltage/ Link to comment Share on other sites More sharing options...
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