So here we are with the second part of the series. If you haven’t read the first part, suggest you do it. Now the answers to the last parts questions.

##### At what RPM will an ideal motor run if I apply 17 volts?

17000 RPM. This is true regardless of what prop I put on the motor because the motor always turns exactly 1000 rpm for every 1 volt applied to it.

##### How much current will an ideal motor draw if a 12×8 prop is attached to it and run on an ideal cell? 2 cells? 3 cells?

According to tables in the article, the current values are 10, 20 and 33 amps respectively. Note that the current goes higher and higher as the voltage increases. These values are mythical and are simply chosen for demonstration purposes.

##### Suppose I had an ideal airplane that was lacking I power but I didn’t want to add cells, what would you suggest I do?

Switch to a bigger prop. Our ideal motor will turn the new prop at the same rpm as the smaller one, which will provide the airplane with more power. The new prop will draw more current than the old prop.

Now onto the next part :

### Current Consumption and Cell Capacity

Today we are going to expand our ideal power system by improving our definition of our ideal cell. Thus far, we have described an ideal cell based on a single characteristic: the fact that it only produces only 1 volt of electricity.

We have also explored how much power is demanded by the motor based on the kind of prop we stick on the shaft and the number of cells we use. But it is at this point that reality rears its ugly head and forces our way into our discussion. Alas, we cannot have all the energy we want forever. Even our ideal cell has a limit to how much energy it can store and, therefore, provide to the motor. This limit is called cell’s capacity.

Let’s fix our ideal cell capacity at a nice round number of 1 amp-hour. This means that our cell can store enough energy that it can produce 1 amp for 1 hour. If we take out more than 1 amp then the cell will run out of energy earlier than 1 hour. For example a 2 amp draw would deplete the cell of energy in half an hour and a 4 amp draw will deplete the cell of energy in 15 minutes.

## Length of time a 1 amp-hour cell can Supply Energy at Various Current Draws

Current | Time |
---|---|

1 amp | 60 minutes |

2 amps | 30 minutes |

3 amps | 20 minutes |

4 amps | 15 minutes |

.... | .... |

20amps | 3 minutes |

### The Duration Equation

Lets express the above as an equation. Our ideal cell makes this easy for us:

##### Duration= 60/Current

The duration is the length of time (in minutes) that a cell can provide energy, given the measured current draw. Note that the above formula is only valid for ideal cells. The real world formula is slightly complicated.

### Adding cells together

In last part, I discussed using more than one cell to power a motor. These cells were assumed to be wired together in a chain, with the positive end of each cell connected to the negative end of the next cell. Series winding adds up the voltage but keeps the capacity same as that of a single cell.

## Voltage and Capacities of cells Connected in Series

No of Cells | Voltage | Capacity |
---|---|---|

1 | 1 Volt | 1 amp-hour |

2 | 2 Volt | 1 amp-hour |

3 | 3 Volt | 1 amp-hour |

4 | 4 Volt | 1 amp-hour |

As you can see in the above table, wiring cells in series does not change the capacity of the battery.

### Computing Duration of Various Power Systems Combination

Now that we know how to compute duration, we can take another look at the power systems presented in 1st part and compare them in more detail.

## Duration of Various Power Systems

No of ideal Cells | Current | Prop | RPM | Power | Duration |
---|---|---|---|---|---|

1 | 1 | 5x5 | 1000 | 1 Watt | 60 minutes |

2 | 2 | 5x5 | 2000 | 4 Watts | 30 minutes |

3 | 3 | 5x5 | 3000 | 9 Watts | 20 minutes |

4 | 6 | 5x5 | 4000 | 24 Watts | 10 minutes |

No of ideal Cells | Current | Prop | RPM | Power | Duration |
---|---|---|---|---|---|

1 | 10 | 12x8 | 1000 | 10 Watt | 6 minutes |

2 | 20 | 12x8 | 2000 | 40 Watts | 3 minutes |

3 | 33 | 12x8 | 3000 | 90 Watts | 2 minutes |

4 | 63 | 12x8 | 4000 | 240 Watts | 1 minute |

In each row of the table, note that the value in the duration column is equal to 60 divided by the value in the current column. The duration starts off at 60 minutes but drops dramatically as current goes up or the prop size is increased. One thing that escapes many newcomers to electric fliers is that adding cells will **decrease the full throttle duration of a power system**. Take a look at the above table and see for yourself how the duration drops as cells are added. This is because the current draw goes up as the voltage increases. The result is that adding a cell will give you more power for less time. Of course, if you have an Electronic speed controller to vary the throttle setting in-flight then this is not an important concern.

### Watt Hours

Since our ideal cell can deliver 1 volt and I amp-hour, and since we know that volts and amps combine to make a watt, we can also describe our ideal cell using the term watt-hours. A watt-hour is a unit that describes how much total power is available from a cell over a given amount of time.

##### Watt-Hours = Volts x Amp-Hours

Naturally, I’ve thought ahead on this and chosen really easy values that make the Watt-Hour rating easy to compute. Our cell is exactly rated at 1 Watt-Hour because it provides 1 volt and has a 1 Amp-hour capacity. If we combine cells into battery through a series connection then we are adding up watt-hours. Let’s jump back to an earlier table and add a Watt-Hour column.

## Voltage and Capacities of cells Connected in Series

No of Cells | Voltage | Capacity | Watt-hours |
---|---|---|---|

1 | 1 Volt | 1 amp-hour | 1 Watt-hour |

2 | 2 Volt | 1 amp-hour | 2 Watt-hours |

3 | 3 Volt | 1 amp-hour | 3 Watt-hours |

4 | 4 Volt | 1 amp-hour | 4 Watt-hours |

Imagine for a moment that our ideal cell supplied two volts. This would double the amount of energy in the cell and thereby double how much power we could get out of it in a hour. This kind of cell would have a watt-hour rating of 2. Like-wise, doubling the capacity of the ideal cell would double the watt-hour rating.

Based on idea of watt-hour, we can start to check our work. Since our perfect motor system has no losses, we can double check the watt-hour rating coming out of the system based on the watt-hour going into the power system. From now on I’ll be abbreviating watt-hours and amp-hours as Wh and Ah respectively.

## Comparison of Watt-hours IN and OUT of a power system

No of Cells | Current | Watt-hours (IN) | Prop | RPM | Power | Duration | Watt-minutes (OUT) | Watt-hours (OUT) |
---|---|---|---|---|---|---|---|---|

1 | 1 | 1 Watt-hour | 5x5 | 1000 | 1 Watt | 60 minutes | 60 | 1 Watt-hour |

2 | 2 | 2 Watt-hours | 5x5 | 2000 | 4 Watts | 30 minutes | 120 | 2 Watt-hours |

3 | 3 | 3 Watt-hours | 5x5 | 3000 | 9 Watts | 20 minutes | 180 | 3 Watt-hours |

4 | 6 | 4 Watt-hours | 5x5 | 4000 | 24 Watts | 10 minutes | 240 | 4 Watt-hours |

No of Cells | Current | Watt-hours (IN) | Prop | RPM | Power | Duration | Watt-minutes (OUT) | Watt-hours (OUT) |
---|---|---|---|---|---|---|---|---|

1 | 10 | 1 Watt-hour | 12x8 | 1000 | 10 Watt | 6 minutes | 60 | 1 Watt-hour |

2 | 20 | 2 Watt-hours | 12x8 | 2000 | 40 Watts | 3 minutes | 120 | 2 Watt-hours |

3 | 30 | 3 Watt-hours | 12x8 | 3000 | 90 Watts | 2 minutes | 180 | 3 Watt-hours |

4 | 60 | 4 Watt-hours | 12x8 | 4000 | 240 Watts | 1 minute | 240 | 4 Watt-hours |

So, referring to table above, we can see that a 3 cell battery pack contains 3 Wh of power. This is because it is composed of 3 cells, each of which contains 1 Wh. This 3 cell pack produces 9 watt for 20 minutes when powering our ideal motor fixed to a 5×5 prop. This is the same as 180 Watt-minutes, or 3 Wh. As you can see, the number of Wh going into the system is the same as the number of Wh coming out of the system. With the 12×8 prop, our 3 cell pack produced 90 watts for 2 minutes. Again, this is 180 Watt-minutes, or 3 Wh.

Let’s summarize our model of an ideal power system:

- The ideal motor spins at 1000 rpm for every volt of energy supplied to it, regardless of load.
- The ideal cell produces exactly 1 volt of energy and has a capacity of 1 Ah.

**And let’s summarize what we’ve learned in this part:**

- The duration of an ideal cell is equal to 60 divided by the current being drawn.
- Capacity is specified in Amp-hrs (Ah). A cell rated at 1 Ah can provide one amp for one hour.
- Ideal cells connected in series to form a battery have their voltages added, but their capacities do not add together.
- The amount of power a cell can produce over time is indicated by its Watt-hour rating. This is the product of the cell’s voltage and Amp-hour rating. Our ideal cell is rated at 1 Watt-hr meaning it can provide 1 watt for 1 hour.
- The no watts going into our ideal system is equal to the number of watts coming out. Likewise the number of watt-hrs going into our perfect motor system is equal to the number of watt-hours coming out.
- A battery can deliver a little power for a long time or a lot of power for a little time. Either way the total amount of energy coming out is fixed by its Wh rating.

### Questions for this part

- What happens to the full throttle duration of a model if I add a cell and change nothing else? Why?
- How long can an ideal cell produce 15 watts of power?
- Suppose I had a cell that provided 2 volts of electricity and had a capacity of 2 Ah. How many watt-hours would that be? How long the cell could produce 20 watts? How long can the cell provide 8 amps of current?
- Suppose you were designing a plane to compete in a distance task. What could you do to the battery to make the plane travel faster?
- What does increasing prop size do to power? Duration ? How can I increase both power and duration of an electric power system?
- I need a power system that provides 300 watts from a battery composed of ideal cells. List two different volt/current combinations to accomplish this and duration they will provide at full throttle?