# Calculating proper specs of DC motor for rig driven on the rope

I'm trying to build a cablecam, a picture that serves only for illustration is:

The DC motor is

• 48V
• 5000W
• 6000rpm
• 6000 oz-in stall torque
• 85% effiencity
• 0.066 Ohms Resistance
• 135Kv (RPM / Volt)
• Reduction ratio with timing belt is 1:2, so speed on 60mm output pulley is 3000rpm.

Battery is Li-Ion

• 48V
• 15Ah
• 70A constant/120A peak for 20sec.

Mechanical properties

• Max load 40 kg (including motor, wires, frame, electronics, pulleys...)
• Friction coefficient is 0.7

I want to know if this 48V DC motor can even start this mass on the rope without burn itself. I would like to know if it could reach speed of 50km/h within 10 seconds or 50 meters of travel on rope.

• Please check whether 6000rpm is rated speed or no load speed. – Huisman Mar 6 at 22:38
• Thx @Huisman for your effort! No load rpm's are 6400, so I assume 6000rpm is rated speed. The weight of the engine is 7.3 pounds, but it is already included in the total weight of 40kg, together with the weight of the frame, electronics, pulley and additional elements. Efficienty of motor is 85%. For this 48V model the manufacturer did not put the curves in the specification, only for 24V model. – littlerock Mar 7 at 7:18

MOTOR REQUIREMENTS

Start-up
The force to overcome the friction equals $$F_s^{max} = \mu_s m g=0.7 \cdot 40\text{ kg} \cdot 9.8\text{ m/s}^2=274.4\text{ N}$$

Required torque of the output pulley to start up equals $$\tau_{pulley} = F_s^{max} \cdot \dfrac{d_{pulley}}{2} = 274.4\text{ N} \cdot 30\text{ mm} = 8.2\text{ Nm}$$

Reduction ratio is 1:2, assuming a efficiency of 90% means the motor should provide a torque that equals $$\tau_{motor} = \tau_{pulley} \frac{1}{i} \frac{1}{\eta} = 8.2\text{ Nm} \cdot \dfrac{1}{2} \cdot \dfrac{1}{0.9}= 4.6\text{ Nm}$$

6000 oz-in stall torque (which is the torque at 0 rad/sec) equals 42.3 Nm which is more than enough to overcome the required 4.6 Nm.

Acceleration
The force to accelerate the load equals $$F_{acc} = ma = 40\text{ kg} \cdot \dfrac{ 50\text{ km/h}}{10\text{ sec}} = 55.6\text{ N}$$

Note that the friction still needs to be overcome, the required force equals $$F_{total}=274.4\text{ N} +55.6\text{ N}= 330\text{ N}$$

Applying the same equations leads to $$\tau_{motor} = 330\text{ N} \cdot 30\text{ mm} \dfrac{1}{2} \dfrac{1}{0.9} = 5.5\text{ Nm}$$

In motion
When the rig is at 50 km/h, the friction force still needs to be overcome.
The required angular velocity of the motor (using the reduction ratio) equals $$\omega_{motor} = \frac{v_{rig}}{d_{pulley}/2} \cdot i= \frac{13.9\text{ m/s}}{30 \text{ mm}} \cdot 2 = 926\text{ rad/s}=8842 \text{ rpm}$$

This is exceeding the motor speed

When changing the reduction/gear ratio to 1:1, the motor torque to overcome friction becomes $$\tau_{motor} = 9.1\text{ Nm}$$ and the torque to accelerate $$\tau_{motor} = 11\text{ Nm}$$.
The required velocity in motion becomes $$\omega_{motor} = 4421 \text{ rpm}$$

NOTE
I used a load of 40 kg, but such a big motor has quite some weight as well. This should be taken into account and recalculation needs to be done.

GRAPH WITH MOTOR CURVES
Some motor manufacturers provide curves with speed, power and efficiency with respect to torque.

Mechanical working points
In this graph the blue curve is relevant for the (mechanical) equations done above. This line represents what speed the motor can achieve for given torque at the rated voltage of 48V.

• At start up, the motor speed is 0 rpm and the required motor torque is 9.1 Nm, so we can draw a point at (9.1 Nm, 0 rpm).
• In motion, the motor speed is 4421 rpm and the required motor torque is 9.1 Nm, so we can draw a point at (9.1 Nm, 4421 rpm).
• During acceleration the required motor torque is 11 Nm and the motor speed increases from 0 rpm to 4421 rpm, so we all these points (forming a line) from (11 Nm, 0 rpm) to (11 Nm, 4421 rpm).

All points are drawn left and below the blue line, which means the motor is able to start up, accelerate and keep the rig in motion.

It seems dull to add the start up and in-motion points, as the acceleration points cover these already. However, some systems need to overcome more start-up forces and/or apply a different acceleration profile.

For dc motors applies:

• the motor torque is proportional to the motor current
• the motor speed is proportional to the motor voltage

Reducing the applied motor voltage shifts the blue line. Controlling a dc motor is "shifting the blue line downwards/to the left" by lowering the motor voltage till it intersects with the points drawn above. (It is possible to overdrive the motor with a higher voltage than 48V, but I wouldn't recommend it. Moreover, as your battery voltage is 48V, it would require addition means)

Note that the battery voltage may lower during use, so, changing the reduction/gear ratio to 4:3, gives more headroom.

Motor current The red curve in the graph is the current for given torque. The current ratings for the motor have been given as:

• no load current = 4.5 A (at zero torque = 0 Nm)
• stall current = 650 A (at stall torque = 42.3 Nm)

So, the relation between torque and current is $$I_{motor} = \tau_{motor} \frac{650\text{ A}-4.5\text{ A}}{42.3\text{ Nm}} + 4.5A$$

(Using the reduction/gear ratio to 1:1) the required torque to accelerate eqauls 11 Nm, so the required current becomes 172 A.
To keep the wig in motion, a torque of 11 Nm is required, requiring a current of 143 A.

**This exceeds the current ratings of the motor controller (and battery) **

Using the former reduction ratio to 2:1 requires 4.6 Nm to keep the wig in motion, so a contineous current of 74 A. This almost satisfies the current rating of the motor controller, but the required speed cannot be achieved as shown above.

SUMMARY

• The motor is able to comply the requirements, especially using a gear ration of about 4:3.
• The motor controller fails to comply as it either limits the maximum torque or the maximum speed.

Note that the friction force is 5 times as big as the acceleration force. So, down-scaling the requirement for acceleration has hardly no use.
The friction force is proportional to the payload as well as the friction coefficent. Improving the latter may yield interesting results.

• I believe there's a typo on your $\tau_{motor}$ equation, which should be 0.030m instead of 0.030mm. – Wasabi Mar 7 at 0:41
• Thanks for cleaning up and pointing to the typo. Has been corrected. – Huisman Mar 7 at 7:37
• With this calculation it turns out that the DC motor will not be a problem, but a battery that can not give more than 120A in a short time. If I understand correctly, the next step is how much torque can produce a given motor at 70A constant, or 120A peak current, and whether that will be enough. Great work @Huisman! – littlerock Mar 7 at 15:25
• Do you know the stall current of the motor? – Huisman Mar 7 at 15:44
• According to the manufacturer: no load amps - 4.5A, amps at 85% efficienty - 55A, stall current - 650A, but the motor controller (and battery) can supply 70A constant, and 120A at the peak, not longer than 20 seconds. – littlerock Mar 7 at 17:15

I think it goes something like this - the energy from the motor is just it's rated power $$P$$ $$\times$$ time $$t$$, which produces kinetic energy (mass $$m$$, velocity $$v$$) in the system: $$P t = \frac{1}{2} m v^2$$ solve for $$t$$, knowing $$P$$ (5000 W), $$m$$ (40 kg) and final $$v$$ (50 km/h) results in the time to reach 50 km/h is about 0.8 s.

It's a simplification so I'm disregarding

• slipping of the drive wheel on the rope - likely to be some initial slippage with a constant power input because the initial acceleration is infinite.
• additional mass of the motor raising the total accelerated mass over OP's 40 kg
• any motor/battery/efficiency related power limitations
• might need a different gearing ratio over what OP has specified (3000 rpm on 60 mm diameter pulley is 33 km/h)
• friction in pulleys, inertia of pulleys

The frictional between the drive pulley and the rope isn't just related to the mass of the cart, it's also related to the tension in the rope.

• The benefit of the energetic approach is that it is way easier to calculate. The draw back of the energy calculation approach is the following: consider having 2 motors: motor A rated 10 mNm and 1000000 rad/sec = 10kW and motor B rated 50 Nm and 100 rad/sec. A lot of engineers would pick motor A because it is the most powerfull. But motor A will fail horribly in OP's application. – Huisman Mar 8 at 10:41
• I dont say the energic calculation approach is wrong, it does give some insgight, but also do calculate the forces etc. – Huisman Mar 8 at 10:43