# FRC team 5584. Est. 2014

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Assumptions:

The totes weigh 7.8 lb (3.5 kg) each.

The containers weigh 8.65 lb (3.9kg) each.

The gantry weighs 22 lb (10kg).

References:

REF#01: FRC “2015 Motor Information” table – see http://:rps01.usfirst.org/frc/manual/2015_motor_information.pdf

REF#02: FRC 2011 Motor Information – see http://www.usfirst.org/uploadedFiles/Robotics_Programs/FRC/Curve_and_Season_Info/2011_Assets/Kit_of_parts/Motor_Curves_Rev_A.pdf

REF#03: Andy Mark CIM Motor Curves – see http://files.andymark.com/CIM-motorcurve.pdf

These references are also available here: REF#01, REF#02 and REF#03

Motor Selection:

If our aim is to lift a single tote box 0.7m in 1 second, what power is required to achieve this?

Work done, force and distance are related to mass and acceleration as follows ...

```Work = Force x Distance
Force = Mass x Acceleration
Power = Work / Time
```

Putting these together ...

```Power = Mass x Acceleration x Distance / Time
```

The mass of a single tote is 3.5 kg. Acceleration due to gravity is 9.81 m/s2. Distance is 0.7 m and the time is 1 second. With these values the power requirement in Watts is ...

```Power = 3.5 x 9.81 x 0.7 / 1 = 24 W
```

Note that this figure represents the minimum power requirement, it neglects losses due to friction and the weight of the gantry.

What if we want to lift more than one tote? Say 4 totes and a container. In this case the mass would be (3.5 x 4) + (3.9 x 1) = 17.9 kg so the power requirement in Watts would be ...

```Power = 17.9 x 9.81 x 0.7 / 1 = 123 W
```

Which of the motors premited in the 2015 FRC season meets this requirement? We can see from the table in REF#01 that the following motors meet or exceed the power requirement ...

• CIM = 337.34 W
• RS-550 = 245.83 W
• RS-775 = 266.79 W
• 9015 = 179.49 W
• BAG = 146.71 W
• Mini CIM = 227.40 W

Lets consider the standard CIM. This has a maximum power of 337.34 W which meets our requirement. If we draw a horizontal line at 123W on the CIM curve in reference REF#02 we can find 2 points on the torque-power curve where power is 123W: These correspond to motor torques of approximately 30 oz-in and 320 oz-in. The better efficiency (and acceptable current) is at 30 oz-in. In fact, the efficiency at 30 oz-in is approximately 65% which is close to the maximum efficiency for this motor. At this point the current is under 20 A so we are well within the specification of the motor (see REF#03) and also under the supply and cable rating (assuming we are using one of the 40 A circuits on the PDP).

If 4 totes plus a container is our maximum load and we configure the transmission to lift in 1 second at 30 oz-in torque then we are using the motor at near peak efficiency, this makes the CIM an ideal choice for this application.

Transmission Configuration:

Ok, so we have selected a motor (CIM) and we have a defined our maximum load as 4 totes plus one container (17.9 kg). How do we set up the transmission for the motor to run at the required torque? The output torque of the transmission (To) is related to the mass and the distance that it is being applied from the axis of the pulley …

```Torque Out = To = Mass x Distance = 17.9 x 0.03 = 0.537 kg-m
```

Lets assume we use belt drives with 60mm diameter (0.03m radius) pulleys to accomplish the lift.

The torque at the motor is the input torque of the transmission (Ti). The input and output torque are related to the gear train or speed ratio (e), the single stage transmission efficiency (n, which is typically 0.95) and the number of stages in the transmission (S) as follows ...

```Ti = To x e / n^S
```

How is the gear train or speed ratio calculated? It is determined by dividing the product of the number of teeth on each of the driving elements by the product of the number of teeth on each of the driven elements.

```Speed Ratio = driving elements / drive elements
```

For example a single stage transmission with a 12 tooth driving gear on the motor axel and a 60 tooth driven gear on the output shaft will have a speed ratio of ...

```Speed ratio = 12 / 60 = 0.2
```

How many stages will our transmission have? There are many options here but for the purposes of calculating the overall transmission efficiency lets assume there are five stages (two in the Andy Mark ToughBox Mini transmission plus a chain drive plus two toothed belt drives, for the purpose of this calculation we ignore the fact that the belt drives may be duplicated via a common axle).

Ok, so lets get back to the torque equation, if we assume a five stage (S=5) transmission with a single stage efficiency of 0.95 (n=0.95). From the curves in REF#02 we know that we need 30 oz-in (0.022 kg-m) at the motor and the calculations above tell us that the output torque will be 0.537 kg-m. We can rearrange the torque equation that we introduced above ...

```Ti = To x e / n^S
```

to ...

```e = n^S x Ti / To = 0.95^5 x 0.022 / 0.537 = 0.031
```

A speed ratio of 0.031 corresponds to a gearing ratio of 1:32.

How might we obtain the overall speed ratio? Another way of writing the speed ratio is as follows ...

```Speed Ratio, e = e1 x e2 x e3 x e4 x e5
```

Two stages of the transmission are toothed belts with driven and driving pulleys of the same size so e4 and e5 are both 1. The first two stages are inside the toughbox mini, if this is configured with ...

• stage 1: 14T driving gear and 50T a driven gear
• stage 2: 16T driving gear and 48T a driven gear

... so e1 = 14/50 = 7/25 and e2 = 16/48 = 1/3. If we require a total speed ratio of 0.031 then the above equation becomes ...

```0.031 = 7/25 x 1/3 x e3 x 1/1 x 1/1
```

Which gives ...

```e3 = 0.031 x (25 x 3 x 1 x 1) / (7 x 1 x 1 x 1) = 0.33
```

The chain drive needs to have a speed ratio of at least 0.34 (or 1:2.9). If we use a 12T driving sprocket and a 60T driven sprocket for the chain stage then we will provide 1:5 which is well above the requirement and will certainly be capable of lifting the four totes, however, selecting a higher gear ratio will result in the lift taking longer. Of course we should repeat the calculations to include additional weight (two extra totes plus the weight of the gantry itself). Note that the required gear ratio with additional weight will need to be lower if the lifting time is to be maintained. The following table shows the consequences of working through the above calculations with different weights:

 LIFTING Weight (kg) Power (W) Ti (kg-m) To (kg-m) e e3 4 totes + container 17.9 123 0.022 0.537 0.031 0.33 (12T/36T) 6 totes + container 24.9 171 0.036 0.747 0.037 0.40 (12T/30T) 6 totes + container + gantry 34.9 240 0.058 1.047 0.043 0.46 (12T/26T)

The full load of 6 totes plus container and gantry will reduce the motor options to just:

• CIM = 337.34 W
• RS-550 = 245.83 W
• RS-775 = 266.79 W

In our case, the mounting options available to us meant that we stuck with the CIM.

Notes:

(a) once we implement the design, the gearing is fixed. The calculations are all on considering the maximum load, they are assuming that we always lift the same number of totes/containers. What will happen if we lift less? What happens if we lift more?

(b) It is assumed that there is no friction in the lifting system. This is clearly not realistic and we should make some allowance for this!

(c) the belts in our solution are duplicated from the common axel (i.e. stages 4 and 5) of the transmission. We have assumed that this has no impact on the transmission calculations, this may not be a valid assumption!