

Michigan
Joined Sep 2005
432 Posts

Discussion
AMPs and other general information
I have been reading a lot of posts on this forum that deal with "what amps am I drawing.." Is there a place (thread, site, etc.) that talks about this in detail? I cannot get it straight in my mind why changing prop size would affect the AMP's being drawn!
I have looked at the RC Power FAQ and it talks a little bit about it but not much. It mentions a C rating (one times the mAH capacity) that is used to determing the charge/discharge rates of a battery  is that the same thing that people are looking at when trying to determine if a power system is overdrawn? If so then where do you find out what C rating your battery has? I guess I am looking for a "AMP's/batteries/motors for dummies" document! :) Thanks 



LI, New York, USA
Joined Mar 2003
25,437 Posts

First the background:
This note is intended to clear up a few terms and concepts around electricity as it applies to electric airplanes. Think of electricity like water. Volts = pressure Amps = flow Volts is like pounds per square inch, psi. Says nothing about how much water is flowing, just how hard it is being pushed. You can have 100 psi with zero water flow. Amps is flow, like gallons per hour. You can have flow at low pressure and you can have flow at high pressure. Amp hours is how much flow can be sustained for how long. It is used as a way of measuring how much electricity is in the battery. Like how many gallons of gas in your tank. It is a capacity number. Says nothing about flow or pressure, it is about capacity. Amps and mili amps? We are just moving the decimal point around. 1 amp (short for ampere)  1000 miliamps (mili means 1/1000) Examples So a 7 cell NIMH or NICD pack provides 8.4V (pressure). The motor will draw electricity from the pack at a certain flow rate, or amps. If you have a have a 650 mili amp hour pack, it can deliver a flow of .650 amps (650 miliamps) for one hour. If you draw it out faster, it doesn't last as long. So your motor might pull 6.5 amps for 1/10 of an hour, or about 6 minutes. A 1100 mah pack has double the capacity of the 650 mah pack, so it should last "about" twice as long. What is C in relation to batteries? C ratings are simply a way of talking about charge and discharge rates for batteries. 1C, = 1 time the rated mah capacity of the battery. So if you charge your 650 mah pack at 1C, you charge it a 650 miliamps, or .650 amps. 1C on a 1100 pack would be 1.1 amps. 2 C on your 1100 pack would be 2.2 amps Motor batteries are often rated in Discharge C and charge C. So a 1100 mah pack (1.1 amp hour) might be rated for 10C discharge, so you can pull 11 amps ( flow ) without damaging the battery. Then it might be rated at 2C charge rate (flow), so you charge it at 2.2 amps (2200 mah) How did I do? Things clearing up? If you have a 500 mah pack  any kind  and it is rated at 16C that means it can deliver 8 amps. If you have a 1000 mah pack  any kind  and it is rated at 8C that means it can deliver 8 amps. If you have a 1000 mah pack  any kind  and it is rated at 12C that means it can deliver 12 amps If you have a 1500 mah pack  any kind  and it is rate at 8C that means it can deliver 12 amps If you have a 1500 mah pack  any kind  and it is rated at 20 C that means it can deliver 30 amps. If you have a 3000 mah pack  any kind  and it is rated at 10 C that means it can deliver 30 amps. So, if you need 12 amps you can use a pack with a higher C rating or a pack with a higher mah rating to get to needed amp deliver level. One last point. Motor batteries vs receiver batteries Some batteries can sustain high discharge rates. Others can not. Those used as transmitter/receiver packs typically are made for low flow/amp rates while those made for motor packs can sustain higher rates. So, having a 600 mah pack does not tell you if it is a motor pack that can put out 6 amps, or if it is a transmitter/receiver pack that would be damaged if you tried to pull power at 6 amps. It is enough to say that they are different. Clearly a motor pack could be used for a transmitter/reciever job, but a transmitter/reciever pack should not generally be used as a motor pack. Basics: http://www.modelaircraft.org/mag/FTGU/Part8/index.html Lithium Batteries http://www.rchobbies.org/lithium_bat...eakthrough.htm New Electric Flyer FAQs http://www.ezonemag.com/pages/faq/a105.shtml 


LI, New York, USA
Joined Mar 2003
25,437 Posts

Now to your question about propellers.
Your electric motor draws a certain amount of energy to do its job, which is to turn the propeller. With no prop attached it draws very little energy. If you put a big prop on the motor it draws a lot of energy. This is similar to pulling a boat trailer behind your car. The car might get 20 mpg normally, but put a boat on a trailer behind the car and mileage will drop off to perhaps 15 mpg because the motor is using more energy just to maintain the same speed and travel the same distance. However as long as the boat and trailer are not too heavy, no real damage occurs, you just use more gas. If you put too big a trailer behind your car, something will break. The motor may fail, the transmission may fail or something else. That is because you are asking the drive train to produce more work, use more energy then it was built to handle. Fuel mileage goes way down and then something breaks. You have over stressed things. Back to your plane. Your electric motor needs to "draw" a certain amount of energy in order to turn a given propeller at a given speed. Let's use a speed 400 motor as an example and let's say you have a 6X5 prop on it. That means the propeller is 6" across and has a pitch of 5" per revolution. Pitch indicates how far the prop would move forward through the air if there was no slippage. As either of these numbers go up, the motor is asked to do more work. Now let's apply some numbers. These are made up numbers for illustration only. Don't assume that these are accurate for your motor in your plane turning your prop. Let's say that, to turn that 6X5 prop your speed 400 motor draws 6 amps of electricity using a battery that delivers 10 volts, just to make the math simple. That would be 60 watts of energy that the motor consumes to turn that prop. (6 amps X 10 Volts) If we go to a larger prop, say 7 inches and keep the pitch the same 5 inches, the draw might go up to 8 amps at 10 volts or 80 watts. Likewise if we went to a 7X6 prop, the draw would go up again, say to 9 amps or 90 watts. In each case we are increasing the amount of work the motor has to do to turn the prop. The harder it works the more electricity it draws. This is also placing an increasing amount of stress on the motor causing it to generate heat and placing more pressure on the bearings. If we push it too far, the motor will be unable to turn the prop fast enough to be useful in flying the plane and/or it will fail from stress, just like the car example above with the trailer that is too big. What we are try to do is to get the best balance of propeller and amp draw so that the motor operates efficiently without being over stressed. Likewise if you have that same speed 400 motor and keep the prop at 6X5 but increase the electric pressure, volts, to 12 volts it will force more amps into the motor. This would be like putting a supercharger on your car's motor which forces more fuel/air mix into the car's engine. It will produce more power so it can do more work. However if we exceed the amount of power it was designed to handle, it will fail. It might not fail right away, but over a very short time it will start to degrade, perform badly and perhaps suddenly fail all together. If we push the voltage up too high or the amp draw too high, we will over stress the motor and damage it. The goal is get a good balance of propeller and power draw. Was that helpful? 


As far away from RCGroups and the AMA as possible!
Joined Aug 2004
6,601 Posts

Wow  Aeajr did a GREAT job in explaining everything. How long did it take you to type all that?!
ElSeeker  If you'd like to play with some numbers, try this web site  http://brantuas.com/ezcalc/dma1.asp 


LI, New York, USA
Joined Mar 2003
25,437 Posts

All batteries have C ratings, however none are as critical as the Lithium packs.
If you over stress a NICD or NIMH pack you will probably damage the battery or degrade it quickly over time. If that happens, you send it to recycling. While it is possible to have a NICD or NIMH pack explode if you GROSSLY abuse it, it is still relatively unusual. If you over stress a Lithium pack, there is a high probabiliy of an explosion or fire, so people are highly focused on the C ratings for the Lithiums. Lithium is also much more sensative to being discharged too far down. If you run a NICD or NIMH pack fully dead, there is a fair chance you can bring it back with a good charger and some careful cycling. If Lithiums get below about 2.5V per pack, they degrade or become useless very fast. So we all focus on the C ratings of Lithiums very closely. 


LI, New York, USA
Joined Mar 2003
25,437 Posts

Quote:
It took a while, but this is a question that comes up often. I thought I would take the time to work out a good answer that is clear and simple. I hope I accomplished that. I now have it saved so, when the next person asks, this is a cut and paste post. I will evolve it over time so I can help other newbie electic flyers understand this topic. If anyone has other resources or links that would be good addons to this writeup on motors/amps/props, I would love to see them. I am always trying to help the new guys. 



Taipei Shek, Taiwan
Joined Aug 2004
458 Posts

The currentvoltagerpm relationship for an electric motor is actually very simple. Assuming first the bearings are frictionless:
Current = (Voltage  rpm/KV)/Resistance Where resistance is the motor's internal resistance from the coils and such. If you plot current vs. rpm, it'll be a straight line intersecting the current axis at (Voltage/Resistance), while intersecting the rpm axis at (Voltage*KV). You can extend this line to both higher and negative rpm's and it'll still be correct. Further more, the torque a motor produces is about proportional to current, so this line also indicates the torque you get. Notice how the torque is zero at rpm=Voltage*KV, and negative beyond? It's easy to see (if you really plot it), that under the same voltage, a motor draws increasingly more current the slower it's allowed to spin, also producing more torque. It's really simple stuff if you know physics and a bit of math. 



Wow, great stuff aeajr you answered a lot of questions I was yet to ask, the other day I was working on gear ratio calculations for my little Estarter but I didn't get to the prop or voltage calulations yet.. If I understand correctly I can sort of think of the prop my third gear (pinion, drive gear, prop) ? Stop me if I'm wrong here but buy gearing up or down and changing the prop you can achieve similar results? Example, my Estarter (stock setup) uses a D gear setup which has a 3.00 ratio (the motor spins three times for every one on the driven shaft ) and the prop it came with is a 1060, but if I wanted to (help me here) gear up / down I could run a E gearbox which has a ratio of 3.40 I could obtain similar results using a 1047 prop? Now comes the hard part, which ratio / prop combo is going to make life easier for my motor? Sorry if I'm way off here but I to am trying to figure out how to choose the right power setup..




Ed, Im planning on doing another "Newbies Guide", this time i was actually planning on calling it "RC Airplanes for Dummies", complete with the exact style and structure of the For Dummies series. Since i have no intentions of selling anything, i can always take down the page if they send the lawyers after me. I checked their website, they dont (yet) have this topic covered. They do have some ridiculous titles though...
Anyways, i  like yourself  wanna put all this stuff in one place for the average beginner, in a fashion thats easy to understand and doesnt get lost in too much techie babble. I especially like the way the For Dummies series includes techie babble in paragraphs titled "Dont Bother Reading This". It puts the info on the page where the reader may absorb some of it, but makes it clear that actually understanding certain things is optional. Since you already have written an enormous amount of stuff, would you be interested in coauthoring? I can edit together a few of your posts to cover entire chapters....im also gonna need critique on what i have written myself. Im still in the planning stages, though i have Paintshopped the nerdy guy icon by adding a beard and ponytail A slow start for sure.... LMK if your interested! Mike 


Michigan
Joined Sep 2005
432 Posts

Quote:
NIMH 8.4V 600mAh (160) (3) (E08) and the other PKZ 1021 NIMH 8.4V 600mAh(160) Do they normally say on the battery itself what the discharge rate is? Or are NIMH's all about the same so you can go by the # cells? 



LI, New York, USA
Joined Mar 2003
25,437 Posts

Quote:
One key point is that each motor typically has a wattage limit either expressed as amps at a given voltage or watts ( voltsXamps). If you exceed the wattage capacity of the motor you will either shorten its life or destroy it. Likewise your wattage targets will tell you what you need to get out of your battery pack. If you need 15 amps at 9.6 volts from your pack and it can only deliver 12 amps without damage either your setup will perform poorly, your batteries will be damaged, or both. This is a VERY simplified expression of a very complex world, but you get the idea. If you visit this link you will see a table for the GWS 350C motor/gearbox setup. They chart amps at volts based on gearing and prop. This is the best practical illustration of the ideas expressed here. http://www.gws.com.tw/english/produc...em/eps350c.htm In most cases the maker of the model or motor will advise on a motor/prop/volts/amps recommendation. Use those as your starting point. I hope this has been helpful. 



LI, New York, USA
Joined Mar 2003
25,437 Posts

Quote:


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