From the Albuquerque Radio Control Club,
Albuquerque NM
Basics of Electric Flight
by Pat Tritle
I really enjoy getting together with clubs
and speaking to the group about the basics of
electric power. However, because there is so much
information that needs to be passed along, it would
be difficult, if not impossible, for those attending
to remember much of the pertinent information. For
that reason, it's better to write up the basic
guidelines so that those who are interested in
getting into electrics would have the information
available for reference at a later date.
Here goes. I'll keep the numbers as simple as
possible to avoid unnecessary confusion.
OK, here's how it all shakes out. The basic power
required to fly an electric model is as follows:
Direct Drive
Systems 60 watts/pound
Gear Drive Systems 50 watts/pound
Mild aerobatic performance 70-80 watts/pound
For all-out aerobatics 100-110 watts/pound
3-D performance 150 watts/pound or more
The above numbers are based on models with wing
loadings from 8-16 oz/square foot. As with gas
models, higher wing loadings require more power
since they must fly faster to support the added
weight. By the same token, a lightly-loaded model
with a wing loading in the 3-5 oz/square foot range
will fly very well at 25 -30 watts/pound.
What's a 'watt'; and where can I get some?
Wattage is the term used in electric flight to
relate the level of power that an electric drive
system will produce. To relate it to terms we're
familiar with, 746 watts = 1 horsepower. To
calculate the wattage delivered by a given system
looks like this: amps x volts = watts. So where do
these numbers come from and how do I know how many
volts and amps are needed to fly a given model?
Okay, let's say you want a mildly aerobatic sport
model with a 14 oz/square foot wing loading that
will weigh in at 2 pounds. We already know that the
power requirement for a model like this is about 70
watts/pound, so we're going to need to generate
about 140 watts. Let's assume that you are going to
use an eight-cell Ni-Cd battery. At 1.2 volts per
cell, eight cells will deliver 9.6 volts. To arrive
at the necessary current draw to achieve 140 watts,
simply divide 140 (watts) by 9.6 (volts) and you
arrive at 14.58 amps.
Now, let's assume that you have a three-cell Li-Poly
battery for the model, which is rated at 11.1 volts.
The formula is the same; 140 (watts) divided by 11.1
(volts) = 12.6 amps. As you can see, as the
available voltage increases, the lower the current
draw needs to be to deliver the necessary wattage.
Now here's something to consider when selecting your
system: the higher the current draw, the shorter the
flight duration on any given battery. Therefore, the
ideal setup would be to use a higher-voltage battery
with lower current draw for maximum duration. On the
downside, when using Ni-Cd and NiMH batteries, as
the cell count goes up, the weight will increase
significantly as well. It works that way with
Lithium too, but Lithium batteries are dramatically
lighter then the old "round" cells.
Okay, let's say we're going to use an 11.1 volt
Li-Poly battery. All we need to do now is select a
motor that will swing enough propeller at 12.6 amps
to fly the model at a top speed of around 40-45 mph
and we're in business. Now that you know the
parameters, visit your local hobby shop and select a
motor that fits that description.
Gear Drive vs. Direct Drive: Why is one better then
the other?
Well, it all depends on the kind of performance
you're looking for. If you're looking to go fast, go
with direct drive. Going fast requires a high-pitch
propeller turning high rpm. The formula to calculate
propeller pitch speed is an easy one; it looks like
this: rpm x pitch (in inches)/1056 = mph.
Let's say that you are turning a 7-6 propeller at
14,000 rpm. 14,000 x 6 = 84,000/1056 = 79.55 mph
Now, let's assume you are setting up a slow,
relaxing park flyer with about a 5 oz/square foot
wing loading. If we swing a 9-7 propeller at about
3,500 rpm, we'd be looking at a top speed of roughly
23 mph. To swing that much propeller with a small,
light drive system, we would use a gear drive unit
at a very low current draw and a small, light
battery.
Again, to make a known comparison, we can relate all
this to riding a 10-speed bicycle. A gear drive
swinging a big propeller is like riding your bike in
low gear. You pedal like mad with little effort, you
don't go very fast, but you can climb steep hills
with ease. The direct drive system could be compared
to riding the bike in high gear. It'll really go
fast, and even though you're pedaling slower, it
requires considerably more effort.
What all this boils down to is "propeller disc
loading.” We all know what wing loading is: it's the
amount of the model's weight that each square foot
of wing must carry. Prop disc-loading works the same
way. A large propeller will be more lightly loaded,
thus delivering more torque then a smaller propeller
turning high rpm. The tradeoff, of course, will be
speed.
One more thing to cover and we'll give you a rest.
Batteries are rated in "voltage" and "amperage.”
Voltage dictates the amount of power the battery
will deliver. The amperage rating dictates for how
long the battery will deliver that power. To relate
that to glow fuel, consider the voltage as nitro
content. High voltage (nitro) means more power. The
amperage is related to the quantity of fuel, or
simply the "size of the tank.”
To figure the size of battery needed, let's go back
to our 140-watt sport airplane. If we're pulling 14
amps from a 1400 mAh (1.4 amp hour) battery, we will
have full power duration of five to six minutes. In
the real world, with proper throttle management,
you'll see flight times of approximately eight
minutes. These are common flight times, even with
liquid-fueled models.
To arrive at that number, divide the battery amp
rating by the current draw: 1.4 (amp hours)/14
(amps) = 0.1. Then take 60 (minutes per amp hour) x
0.1 = 6 minutes. Now, to double the duration, you
must either cut the current draw in half (to 7
amps), or double the battery size (to 2800 mAh or
2.8 amp hours)—again we see tradeoffs. To reduce the
current draw, we can use a larger, higher-pitch
propeller turning slower with very little weight
penalty. If we double the size of the battery
capacity, the weight penalty is quite high unless we
go over to the new Lithium batteries in which we
will discover we have benefited from a tremendous
weight reduction, but at a higher price then
conventional batteries.
Okay, I promise I'll quit before we all end up in
"system overload.” Once again, there's a tremendous
amount of information here for a newcomer to
electrics to digest, so let's do this: if you have
specific questions about setting up an electric
model, please feel free to drop me a line and I'll
do what I can to steer you in the right direction.
For now, I’ll offer up one last piece of advice. To
get started, work with a known good design, and use
the recommended equipment that has been proven to
work. Talk to the people who are successful and copy
what they're doing. The one thing I do know about
modelers is that they are always willing to share
their knowledge with those interested in what they
are doing.
Contact Pat at
patscustommodels@aol.com
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