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Motors: a System Approach Kurt Heinzmann DEKA Research & Development Corp. January 2007.

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Presentation on theme: "Motors: a System Approach Kurt Heinzmann DEKA Research & Development Corp. January 2007."— Presentation transcript:

1 Motors: a System Approach Kurt Heinzmann DEKA Research & Development Corp. January 2007

2 General Topics Example problems Problem formulation and analysis Manufacturers' torque curves and specification sheets Temperature rise Power loss in battery, wires and other components Gear ratio Review of motors from a previous Kit of Parts

3 Background Energy Power Power loss Analysis Test

4 Energy Work is energy. Example: “effort” times “displacement” –Force is effort –Distance is displacement Power Power is how fast work gets done. Example: “effort” times “speed”

5 Power Power is a measure of how fast work gets done. POWER = EFFORT x FLOW (speed) “EFFORT” –force –torque –pressure –voltage –thinking “FLOW” –travel speed –rotating speed –flow of fluid –flow of electrons –doing

6 Power Loss in the Mechanism Some power from the motor is lost due to friction in the mechanism –Gears, belts, cables –Bearings, guides –Tires, balls, or other deformable items –Damage –Contamination Power loss is heat

7 Power required at the motor Power at the motor = power required at the point of use + power lost in the mechanism Power loss is heat

8 Power loss in the motor Power is lost in the motor due to friction, damping, and electrical resistance Power loss is heat. Overloading will cause excessive temperature rise. Use appropriate gear ratio.

9 Analysis Example problems Important motor parameters Motor model revised to include other losses (wires, battery, switches, fuses, etc.) Gear ratio

10 Basic Theory Torque is rotating EFFORT, speed is rotating FLOW –Torque = force x radius Voltage is electrical EFFORT, current is FLOW of electrons Power = EFFORT x FLOW –Mechanical power P(mech) = torque x speed –Electrical power P(elec) = voltage x current

11 Units, Conversions International System (SI) of units Prefixes: m = milli- = one thousandth (mm, mNm) k = kilo- = one thousand (km, kW)

12 Why use SI units? Fewer mistakes than when using U.S. Customary units A motor converts electrical power to mechanical power. –If we express electrical power and mechanical power in the same units (watts), we know what’s happening at both ends of the motor, and inside it. Many are named after famous scientists Advice: Convert each parameter to SI units before doing any other calculation. Consolation: you can always convert back to US customary units.

13 Problem 1 Accelerate to a speed

14 Problem 1 Mass: m = 150 lb. = 68 kg Speed: v = 6 ft./s = 1.8 m/s Acceleration: a = 1.8 m/s per second = 1.8 m/s 2 Force = m x a = 68 kg x 1.8 m/s 2 = 122 N Force from each wheel: F = 122 N / 2 = 61 N Power: P = F x v = 61 N x 1.8 m/s = 110 W

15 Problem 2 Lift a weight a distance within a time

16 Problem 2 Gravitational constant: g = 9.8 m/s 2 Weight: W = 14 lb. = 61 N Force: F = W = 61 N Height: h = 6 ft. = 1.8 m Time: t = 4 s Speed: v = 1.8 m/ 4 s = 0.45 m/s Power: P = F x v = 61 N x 0.45 m/s = 28 W

17 Basic Motor Theory

18 Electrical Components

19 Basic Motor Theory

20 Basic Motor Theory Important motor parameters Applied voltage ( V ) Stall torque (  stall ) Stall current ( i stall ) Free speed (  free ) Resistance ( R )

21 Fisher-Price Motor

22 From data sheet:  stall = 0.65 Nm i stall = 148 A  free = 2513 rad/s Calculate: Resistance R = 12 V /148 A = 0.081  Fisher-Price Motor (2005) Stall torque Stall current Free speed Reference voltage V = 12 V

23 Fisher-Price Motor – Current (For detailed analysis, see " Getting the Most From Your Motors" by Kurt Heinzmann, 2006)

24 Fisher-Price Motor - Speed

25 Fisher-Price Motor - Power output

26 Fisher-Price Motor - Input Power

27 Fisher-Price Motor - Power loss

28 Fisher-Price Motor - Efficiency

29 Motor performance based on data sheet Peak power occurs when torque =  stall / 2, and when speed =  free / 2

30 Real World: Power loss 14 AWG wire: 3.0 m  /ft. 12 AWG wire: 1.9 m  /ft. 10 AWG wire: 1.2 m  /ft. 6 AWG wire: 0.5 m  /ft. (Copper at 65 °C)

31 Resistance of electrical system components

32 Simplified electrical system model

33 System model Additional resistance reduces stall torque proportionally. Divide the stall torque on the torque/speed diagram by the factor R system /R motor(nominal) Fisher-Price:  stall = 0.65 Nm/2.3 = 0.28 Nm

34 Performance of the system compared with motor performance based on data sheet

35 CIM motor (also known as Chiaphua and Atwood)

36 Stall torque  stall = 347 oz-in = 2.45 Nm Free speed  free = 5342 rpm = 560 rad/s Free current i free = 2.4 A R system /R motor(nominal) = 2.1 Stall current i stall = 114 A CIM motor data and curves

37 Comparison of power available from Fisher-Price Motor and CIM motor

38 Mechanical Components Gear ratio N g =  in /  out Gear efficiency η g = P out /P in  out =  in / N g ;  out = η g x N g x  in

39 "Gear" ratio: Mechanical power transmission efficiency is important Spur gears: 90% per pair Worm and gear: 10%-60% Nut on a screw: 10%-60% Twist cables: 30%-90% Chain: 85%-95% Wire rope (cables): up to 98% Rack and pinion 50%-80%

40 System with gearbox

41 Gear ratio example Fisher-Price motor with gear reduction Given: Gear ratio N g = 4.6:1 Gear efficiency η g = 90% Calculate: Output torque  out = η g x N g x  in = 4.14 x  in Output speed  out =  motor / N g = 0.217 x  motor

42 Is the little motor/gearbox combination the same as the big motor?

43 The big (CIM) motor will not heat up as fast as the small motor, because it contains more material.

44 Problem 1 ( v = 1.8 m/s; F = 61 N) Motor speed:  motor =  free /2 = 559 rad/s/2 = 280 rad/s We wish to try 8" wheels: R wheel = 4" = 0.1 m Wheel speed:  motor = v / R wheel = (1.8 m/s)/(0.1 m) = 18 rad/s Gear ratio: N g =  motor /  wheel = ( 280 rad/s)/(18 rad/s) = 16 Check torque and propulsion force: Usual limit per stage is 5:1 - need two stages. Gear efficiency: η g = 0.9 x 0.9 = 0.81 Wheel torque:  wheel = η g x N g x  stall /2 = 0.81 x 16 x 1.2 Nm/2 = 7.8 Nm Force: F =  wheel /R wheel = (7.8 Nm)/(0.1 m) = 78 N (OK)

45 Just right

46 Problem 2 ( v = 0.45 m/s; F = 61 N) We wish to try a screw with Fisher-Price motor. Screw speed = motor speed:  screw =  free / 2 = 2513 rad/s/2 = 1256 rad/s  screw =(1256 rad/s)/(2π rad/revolution) = 200 rev./s Screw pitch: p = v/  screw = (0.45 m/s)/(200 rev./s) = 0.00225 m/rev. = 0.00036 m/rad (11 threads per inch). Check torque and force: Assume screw efficiency = 20% Torque:  screw =  motor =  stall / 2 = 0.28 Nm/2 = 0.14 Nm Force: F = η g x  screw / p = (0.2 x 0.14 Nm)/(0.00036 m/rad) = 78 N (OK)

47 Summary of motors in the 2005 Kit of Parts sorted by peak output power

48 Comparison of motors in the 2005 Kit of Parts

49 Keep batteries charged. Delivered capacity was only one third of rated capacity.

50 Conclusion Proper motor selection, good wiring, an appropriate gear ratio, aligned mechanical components, and a full battery will keep you alive in the heat of the battle. Power loss is often a significant fraction of the power used to do work. Include all losses in analysis. Analyze, but test, too! Have fun


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