EUCMania Posted December 8, 2016 Share Posted December 8, 2016 Here I would like to collect some of my understandings of the math and phys related to EUC. If you find mistakes, please point out. Notations: V = max speed, v = speed. W = max power of the motor, w = power. F = force the motor exerts at the edge of tire when stalled = max force. f = force the motor exerts at the edge of tire. R = radius of the wheel. L = length of the pedal from the front to the center of pedal connector. M = mass of the rider, m = mass of the EUC. Example: Msuper V3's data is as follows: V=54km/h = 15 m/s, W = 1500, R = 9in = 0.2286 meter, L = 10cm = 0.1 meter. m=20 kg. ---------------------------------------------------------------------------------- Q. How to determine the stall force F? Where is the max power W achieved? Ans: The max power is achieved at v= V/2. The F = 4W / V. and f = F - F / V * v, assuming the (v, f) graph is straight as usual electric motors. Proof: We know that w = fv. The (v, f) graph of an electric motor is a straight line joining the points ( v, f) = (0, F) to (V, 0). The max w = fv occurs when $v=V / 2 and hence f = F/2. Therefore W = FV / 4. From here F = 4W / V follows. endOfProof. _________________________________________________ Example: For MSuper v3, the stall force is 4*W / V = 4*1500 / 15 = 400 N = 40 kg. The force available to drive forward at speed speed V / 2 = 27 km/h is 200 N = 20kg. Example:: If MSper 3 has a mid speed version with max speed set at 40 km / h = 11.11, and I would like to drive at speed 25 km / h. Does this MS version provide more or less driving force than the HS version? Ans: The F for MS version is F = 4W / V = 4*1500 / 11.11 = 540 N = 54 kg. The driving force at the speed v= 25 km / h = 6.944 m/s is f = F( 1- v/V) = 54( 1 - 25/40) = 20.25 kg. Thus, this MS version offers LESS driving force and hence torque than the HS version when driving at speed v > 25 km / h. Because safety at high speed is more important than at low speed, I should take HS version if I drive at speed >= 25 km / h. ------------------------------------------------------------------------------------ Question: What is the max incline I can climb, assuming the pedal is as long as I want? Ans: theta = arcsin( ( F / m+M)g ). Example: For Msuper v3, assuming my weight is 75 kg, the max climbing angle is arcsin( 40/95 ) = 25 degree. If I climb steeper angle than 25 degree , the motor will stall and the control board can burn. I remember someone posted on this forum that he burnt boards twice by climbing a 27 degree incline. -------------------------------------------------------------------------------------- Question: But my pedal's effort length is only 10 cm long, what is the max incline I can drive by pushing the pedal only? Ans: theta = arcsin( ML / (m+M) / R ). Proof: the torque of my push, MgL, should equal to the torque the mortor put out to drive our weight up = R*(m+M)g*sin( theta). Thus the result. Example: For my Msuper V3, my max climbing angle by pushing the pedal is theta = arcsin( 75*0.1 / 95 / 0.2286) = 20 degree. However, this is still too optimistic because I cannot put all my weight at the tip of the pedal. More likely, I can put my weight at L = 8 cm. Then the angle theta = arcsin( 75*0.08 / 95 / 0.2286) = 16 degree. Therefore, to climb an angle > 16 degree, I have to squeeze the calf pads and lean forward to apply more torque. As long as the angle is < 25 degree, I should be able to climb up. ------------------------------------------------------------------------------------- Question: (Over lean prevention ) I understand that at higher speed, if I put all my weight on the tip of the pedal, the pedal may "soften down" to let me fall forward. So the question is at speed v, where should I put my center of weight over the pedal so that overlean does not happen? Ans: your center of gravity on your feet should not be more than l = RF(1-v / V ) / M / g in front of the center of pedal connector. Proof: The torque the motor can provide is f*R which should = the torque you demand = l*Mg where l is the distance from the center of the pedal connector to the center of gravity your feet put on the pedal. Thus, l = f*R / M / g = ... Example: For my Msuper v3, at speed v = V / 2, the l = 0.2286*400*( 1 - 1/2 ) /75 / 9.81 = 0.0621 m = 6.21 cm. If I lean so much that the center of weight on my feet is > 6.21 cm forward, I will exceed the nominal power of the motor. I must tilt back a bit to stay safe. The EUC should be designed to sound alarm in this situation. One way to prevent overlean at this speed is to move my feet back a bit so that it is harder to exceed the threshold l. ------------------------------------------------------------------------------------- Question: What is the minimum breaking to stop time when I drive at speed v0, assuming the pedal is long enough in the backward direction? Ans: T = - V*( m+M) / F * ln( 1 - v0 / V ). Proof. If the pedal is long enough enough in the backward direction, and assume I can apply as much weight in the backward direction as the EUC can support. Newton's second law states (m+M) dv / dt = - f. Solve it with the initial condition v(0) = v0, we get the above formula. In real situation the pedal is not long enough so I cannot force too much backward to avoid falling from the wheel. Then the breaking time should be longer. Example: For my Msuper v3, traveling at speed v0 = V / 2 = 27km/h, and M = 75 kg, the breaking to stop time is T = 15* ( 20 + 75 ) / 400 *ln(2) = 2.47 seconds. If traveling at speed v0 = V / 4 = 13.5 km/h and M = 75 kg, the breaking to stop time is T = 15* ( 20 + 75 ) / 400 *ln(4/3) = 1 second, about <2 meters distance In real situation, probably I have to add 1 or 2 more seconds to it. I must slow down to speed v < V / 4 when approaching slow moving object about 2+ 3 meters away. ---------------------------------------------------------------------------------- Link to comment Share on other sites More sharing options...

zlymex Posted December 8, 2016 Share Posted December 8, 2016 It is a very good attempt, the math and physics are all fine by me. However, the definition and realization of the EUC may not be that ideal. 1. For Gotway V3, 1500W is the max. continuous output power beyond which the motor will be over heated(at a give set of parameters). Therefore, for a short period of time, the max. power can be output is greater. The peak power specified by Gotway is >3000W 2. By the test of the Russian site, the peak power reached to 4600W. Although that is the 84V version, but the motor is the same as the 67.2V version. In another word, the peak power is very much dependant on the battery voltage. 3. Because the internal armature resistance is very small, the stall condition will never be reached. Instead, there are current sensors inside and will limit the max. armature current(equivalent to limit the torque). Therefore, the torque-speed curve is not a straight line anymore as those red lines below, it becomes a two segment-line-shaped as the green. Link to comment Share on other sites More sharing options...

zlymex Posted December 8, 2016 Share Posted December 8, 2016 10 hours ago, EUCMania said: I remember someone posted on this forum that he burnt boards twice by climbing a 27 degree incline. Yes, that was me. However, that happened not because of the stall of the motor, the stall current would be >500A because the armature resistance <0.1 Ohm. It was not because of the current limit either. IMO it was because of an internal bug that exists in the old version of V3(before 1st Oct.). I had the new version now but I dare not try that 27 degree slope again. Link to comment Share on other sites More sharing options...

The Fat Unicyclist Posted December 9, 2016 Share Posted December 9, 2016 I like shiny things! Link to comment Share on other sites More sharing options...

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