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Watts volts and amps? |
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e.g. 1.2 volts x 7 cells = 8.4 volts
For LiPo 3S means 3.7 volts x 3 cells = 11.1 volts.
Volts is a measure of the electromotive force or electrical pressure that you battery is capable of producing. The more volts you have (pressure) the more work you can do with it. To test this out, take a standard motor (say, a Speed 400) and attach a 7.2V battery to it and note the sound it makes. Then attach a 12V battery and notice how it goes much harder. If you measure the amps, you will see that they go up slightly with the extra Volts, but not much because there is no load.
Amps. The result of this pressure is current flow – connect cable to your battery, connect it to a motor and the voltage rushes out of your battery. The rate at which the voltage rushes out is measured in amperes (AMPS) or milli-amperes (mA), which are one thousandth of an AMP. The higher the AMPS figure the more current there is flowing and your battery will be empty sooner. It's a bit like emptying the petrol tank in your car - the faster you take out the fuel, the sooner it will be empty.
The capacity of a battery (which is your tank of fuel) is measured in milli-amp/hours (mAh) or amp/hours (Ah). If you have a 1200 mAh (or 1.2 Ah) battery, and you draw 1.2A of current from the battery, it will take one hour to go flat. If you double the current consumption to 2.4A, it will be flat in 30 minutes. 4.8A and it will be flat in 15 minutes, and so on.
In LiPo terminology, the capacity of your pack is named "C". So, in the first example with a current draw of 1.2A for a 1.2 Ah battery, the current draw is 1C. Double that to 2.4A and it is 2C, and so on.
To increase the current draw (increase the amps), you increase the load on the motor. The amp draw is governed by the load. Fit a bigger prop and your amps go up, fit a smaller prop and the amps go down (all else being equal).
If you want a longer motor run, without changing your prop or number of cells, fit a larger capacity battery pack. That's like fitting a bigger tank!
Resistance. Current is restricted in its flow by any resistance in the circuit – thin wire, poor connectors, long leads, bad solder joints, thin battery straps - all cause resistance, and this resistance is measured in ohms. To make the most of your electric flight installation, pay attention to all these details to improve the efficiency of your set-up.
Power - or Watts. Now let's get to the power equation. Power from and electric motor is actually a combination of volts and amps. Multiply voltage (volts) and current flow (amps) and you get WATTS which is a measure of either power or power consumption. 746 Watts equals one horsepower!
You can actually increase the voltage that you apply to a motor, and reduce the load (fit a smaller prop) and have less power than you started with. You can have high voltage and low amps, or high amps and low voltage and the power can be the same! When working on a power set-up, you need to consider both Volts and Amps.
To make a case study, take the figures for a GEIST 30/6 running on a battery of 7 cells with a 8 x 6 prop. Firstly, we call 7 cells 7 volts – not 8.4 volts – because as soon as a load is dropped across the battery the voltage drops to about 1 volt per cell.
So – 7 volts x 41.5 AMPS (measured) = 290 WATTS power consumption.
This means that the power consumption of this 300 WATT motor is 290 WATTS which is close to its limit. But the power output is somewhat lower than this figure.
Let us take another case:
An UNGER running on 7 cells with a 9 x 6 prop fitted draws 20.7 AMPS
Therefore 7 VOLTS x 20.7 AMPS = 144.9 WATTS power consumption.
But if you study Unger’s figures you will see that this motor is only putting out 106 WATTS which is why the motor is running at 73% efficiency.
Let us rehearse that:
7 VOLTS x 20 AMPS = 144.9 WATTS in, but only 106 WATTS out, means that the efficiency of the motor with this battery/prop combination is 73%.
Unger also states that this motor is putting out 0.129 Newton/metres which is also a measure of power. The Newton is the absolute unit of force in the metre-kilogram-second system of physical units. It is defined as that force which, if applied to a free body having a mass of one kilogram, would give the body an acceleration of one metre per second per second.
Now you see it all – don’t you?! (I just put that last bit in to confuse you properly!!)