Totally lost.
05-27-2010, 07:27 PM
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Totally lost.
Hello!
Long story short: Electric model, I have a motor and ElectronicSpeed Control picked out, specs seem to work (correct me if i'm wrong!), and need a battery!
What size battery would work with this?Number of cells, burst and continuous current, etc.
MOTOR:
http://leadershobby.com/product.asp?ID=9394001110009
Long story short: Electric model, I have a motor and ElectronicSpeed Control picked out, specs seem to work (correct me if i'm wrong!), and need a battery!
What size battery would work with this?Number of cells, burst and continuous current, etc.
MOTOR:
http://leadershobby.com/product.asp?ID=9394001110009
Rotational Speed : 1700 (kv) RPM/V
Continuous Current : 40A
Max. Current : 50A
Input Voltage : 6 - 11.1V
Max. Efficiency : 98%
No Load Current : 0.6A
Power : 560W
ELECTRONICSPEEDCONTROL:
http://leadershobby.com/product.asp?ID=9394001220250
Input voltage: 5.5V-22V (support 2-5S lithium batteries)
Continue current: 45A
Burst current: 55A (within 10 seconds)
BEC output: 3A/5V (linear)
Thank you for taking the time to read all of this!
Graeme
05-28-2010, 08:31 AM
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RE: Totally lost.
Graeme: actually you have almost all the info you need in the specs. Your motor is limited to 40 amps continuous, so you want a LiPO with enough capacity(C) to handle that. Also the max input voltage is three cells (11.1)There are several ways to accomplish this:a three cell 10C pack of 4000mah C, a 15C pack of approximately 2500 mah C, a 20C pack of 200mah C (there are cells with as much as 45C discharge rate, but they are very high end and I suspect from your questions that you're not ready for them).
You will have to watch prop size to stay within the limits of the ESC and motor. In my opinion your ESC is minimal for your motor. You have about 10% "overhead) (the difference between the current ratings of the motor and those of the ESC). I personally like 20%.
If you do not fly at full throttle all the time. You should be fine. Otherwise I would prop the motor for 35amps.
One more thing: a three cell LiPo may be rated at 11.1 v which is the nominal rating or at 12.6v which is the fully charged voltage. This is just a matter of nomenclature choice; they are the same rating cells.
Keep asking questions. The only dumb one is the one that's not asked.
Walt
You will have to watch prop size to stay within the limits of the ESC and motor. In my opinion your ESC is minimal for your motor. You have about 10% "overhead) (the difference between the current ratings of the motor and those of the ESC). I personally like 20%.
If you do not fly at full throttle all the time. You should be fine. Otherwise I would prop the motor for 35amps.
One more thing: a three cell LiPo may be rated at 11.1 v which is the nominal rating or at 12.6v which is the fully charged voltage. This is just a matter of nomenclature choice; they are the same rating cells.
Keep asking questions. The only dumb one is the one that's not asked.
Walt
05-28-2010, 08:34 AM
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RE: Totally lost.
Oops: typo: a 20C pack would have to be 2000 mah not 200. Three things affect your choice:cost, space and weight. Run time is also a factor, but not the most important.
WT
WT
05-28-2010, 05:57 PM
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RE: Totally lost.
Thanks for the prompt feedback!
If the motor is rated at 11.1 V, would a fully charged three cell pack burn it out?
Also, what kind of run times will I get out of the different sized batteries?
Thanks again,
Graeme
If the motor is rated at 11.1 V, would a fully charged three cell pack burn it out?
Also, what kind of run times will I get out of the different sized batteries?
Thanks again,
Graeme
05-28-2010, 07:06 PM
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RE: Totally lost.
No. Not a problem. By 11.1 they are referring to a 3S battery. Motors are actually quite tolerant of voltage. It is excess current that can do them in.
05-28-2010, 07:52 PM
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RE: Totally lost.
Run times are notoriously unpredictable. The quality of your components (in the example which follows below, I assume 100% efficiency of all components which ain't gonna happen) and your flying style will make a huge difference. Take the time to learn how to use drivecalc which is a free online predictor program. As you progress in this aspect of the sport you will want to make a lot of use of it.
I'd give you the formula, but I always screw it up. Basically you firs have to get everything into the same size units. The numbers can get crazy if you reduce them to mah, so use amps. I'll try to give you an example based on your system. Let's take a 20C 2000mah Lipo. That converts to 4 amps. which means that your pack is expected to deliver 4 amps for one hour. Your motor eats 40 amps/hr. So 4 divided by 40 gives you .1 hour. .1 of 60 is 6 minutes. Doesn't sound like much, but remember, not many people fly at full wide open throttle all the time (at least they shouldn't if they fly electric). Also you power system unloads a little bit in the air (not as much as so think) so the amp draw goes down. Amp draw does not decrease proportionally to throttle reduction (you can see this on drivecalc). A 25% cut back in throttle may reduce amp draw by 50% (I don't know of a formula for this, but there must be one). So if you're drawing 20 amps instead of 40 your flight time will double - 12 minutes is pretty good. The better you are at throttle control the longer your duration. The percentages I've stated are for illustration only your current flow may vary (-:
Walt
PS fmw's post is dead on.
PPS like I said I usually crew up the numbers and I did the first time through. Fortunately, I caught my mistake and the correct numbers are up there now. Please note the caveat in my into though. Figure 80% efficiency and you'll always be safe.
WT
I'd give you the formula, but I always screw it up. Basically you firs have to get everything into the same size units. The numbers can get crazy if you reduce them to mah, so use amps. I'll try to give you an example based on your system. Let's take a 20C 2000mah Lipo. That converts to 4 amps. which means that your pack is expected to deliver 4 amps for one hour. Your motor eats 40 amps/hr. So 4 divided by 40 gives you .1 hour. .1 of 60 is 6 minutes. Doesn't sound like much, but remember, not many people fly at full wide open throttle all the time (at least they shouldn't if they fly electric). Also you power system unloads a little bit in the air (not as much as so think) so the amp draw goes down. Amp draw does not decrease proportionally to throttle reduction (you can see this on drivecalc). A 25% cut back in throttle may reduce amp draw by 50% (I don't know of a formula for this, but there must be one). So if you're drawing 20 amps instead of 40 your flight time will double - 12 minutes is pretty good. The better you are at throttle control the longer your duration. The percentages I've stated are for illustration only your current flow may vary (-:
Walt
PS fmw's post is dead on.
PPS like I said I usually crew up the numbers and I did the first time through. Fortunately, I caught my mistake and the correct numbers are up there now. Please note the caveat in my into though. Figure 80% efficiency and you'll always be safe.
WT
05-29-2010, 06:29 AM
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RE: Totally lost.
Personally, I don't see much value in trying to predict run times. As Graeme says, there are lots of variables. You could get a best case guesstimate by working with the MAH figure of the batter. It means milliamp hours and provides the number of milliamps of current the battery will provide in one hour. If we were to take a 2000 mah battery, for example, that is 2 amp hours (a milliamp is 1/1000th of an amp.) That means your battery might be able to provide 2 amps of current for an hour or 20 amps of current for 6 minutes (1/10 of an hour.) I think the result is fairly meaningless.
What you need to do is to time your flights. Start at 5 minutes, for example, and then read the battery voltage at the end of the flight. That will give you a real life measure of how much battery you use in 5 minutes. You may have to cut that back or use a battery with a larger amp/hour rating. Or you may be able to extend the time. Using your own experience and measuring the results is the only meaningful guide. Too many variables otherwise.
My own practice is to own and use a lot of batteries. i go to the field with about 4 batteries for every plane I intend to fly. I don't go past 6 minutes personally on each flight. At my age, concentrating for 6 minutes is a challenge in itself.
I time the flight. When 6 minutes have passed I land and change batteries or planes. At the end of the day I take the batteries home and charge them.
Another option would be to charge batteries at the field using your car battery or whatever facilities the club has. Lots of modelers do that.
What you need to do is to time your flights. Start at 5 minutes, for example, and then read the battery voltage at the end of the flight. That will give you a real life measure of how much battery you use in 5 minutes. You may have to cut that back or use a battery with a larger amp/hour rating. Or you may be able to extend the time. Using your own experience and measuring the results is the only meaningful guide. Too many variables otherwise.
My own practice is to own and use a lot of batteries. i go to the field with about 4 batteries for every plane I intend to fly. I don't go past 6 minutes personally on each flight. At my age, concentrating for 6 minutes is a challenge in itself.
I time the flight. When 6 minutes have passed I land and change batteries or planes. At the end of the day I take the batteries home and charge them. Another option would be to charge batteries at the field using your car battery or whatever facilities the club has. Lots of modelers do that.
05-29-2010, 12:38 PM
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RE: Totally lost.
Formula for Run Time
run time in minutes = (Battery Capacity in Ah/Motor Current in A) x 60
To get the battery capacity in Ah, divide the battery's mAh by 1000.
This is just a rough guide, and it requires that you know approximately how much current your power system draws. A wattmeter will tell you.
Here are some examples for a power system that draws 40A of current.
Example 1:
2500 mAh battery
power system draws 40A
2500 mAh = 2.5 Ah
2.5/40 x 60 = 3.75 minutes
Example 2:
4000 mAh battery
power system draws 40A
4000 mAh = 4.0 Ah
4.0/40 x 60 = 6 minutes
Example 3:
6000 mAh battery
power system draws 40A
6000 mAh = 6.0 Ah
6.0/40 x 60 = 9 minutes
These examples assume your power system is drawing 40A at all times during flight. In reality, you may not be drawing so much current all of the time, so actual flight times may be longer. For a given motor, the actual current drawn by your power system will depend on the size of the prop. A large prop will cause the power system to draw more current than will a small prop.
Unless you are copying a power system that has been used successfully by others, it's a good idea to measure the current with a wattmeter. Otherwise, you are merely guessing, which can be expensive if it results in a damaged motor or ESC.
On another topic, the 11.1V is the so-called "nominal" voltage of a 3-cell lipo. The actual voltage of a fully charged lipo is 4.2V per cell. So a fully charged 3-cell lipo will measure 12.6V with no load. Under load, the voltage will drop to a value close to the nominal 11.1V, often a bit less.
- Jeff
run time in minutes = (Battery Capacity in Ah/Motor Current in A) x 60
To get the battery capacity in Ah, divide the battery's mAh by 1000.
This is just a rough guide, and it requires that you know approximately how much current your power system draws. A wattmeter will tell you.
Here are some examples for a power system that draws 40A of current.
Example 1:
2500 mAh battery
power system draws 40A
2500 mAh = 2.5 Ah
2.5/40 x 60 = 3.75 minutes
Example 2:
4000 mAh battery
power system draws 40A
4000 mAh = 4.0 Ah
4.0/40 x 60 = 6 minutes
Example 3:
6000 mAh battery
power system draws 40A
6000 mAh = 6.0 Ah
6.0/40 x 60 = 9 minutes
These examples assume your power system is drawing 40A at all times during flight. In reality, you may not be drawing so much current all of the time, so actual flight times may be longer. For a given motor, the actual current drawn by your power system will depend on the size of the prop. A large prop will cause the power system to draw more current than will a small prop.
Unless you are copying a power system that has been used successfully by others, it's a good idea to measure the current with a wattmeter. Otherwise, you are merely guessing, which can be expensive if it results in a damaged motor or ESC.
On another topic, the 11.1V is the so-called "nominal" voltage of a 3-cell lipo. The actual voltage of a fully charged lipo is 4.2V per cell. So a fully charged 3-cell lipo will measure 12.6V with no load. Under load, the voltage will drop to a value close to the nominal 11.1V, often a bit less.
- Jeff
06-18-2010, 05:21 PM
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RE: Totally lost.
Dakrat -
The formula is not for any particular plane. The formula works for ANY plane, ANY motor, and ANY prop.
If you know your power system's current requirement (measure it with a wattmeter) and the capacity of your battery, you can use the formula.
A small plane with a low current requirement could use a smaller battery and still have a decent flight time. A big plane would need a bigger battery for the same flight time.
Small plane example:
1500 mAh battery
power system draws 12A
1500 mAh = 1.5 Ah
1.5/12 x 60 = 7.5 minutes
Big plane example:
8000 mAh battery
power system draws 64A
8000 mAh = 8.0 Ah
8.0/64 x 60 = 7.5 minutes
As you can see, the small plane drawing 12A will need a 1500 mAh battery to fly for 7.5 minutes. The big plane drawing 64A will need a 8000 mAh battery to fly for 7.5 minutes.
- Jeff
The formula is not for any particular plane. The formula works for ANY plane, ANY motor, and ANY prop.
If you know your power system's current requirement (measure it with a wattmeter) and the capacity of your battery, you can use the formula.
A small plane with a low current requirement could use a smaller battery and still have a decent flight time. A big plane would need a bigger battery for the same flight time.
Small plane example:
1500 mAh battery
power system draws 12A
1500 mAh = 1.5 Ah
1.5/12 x 60 = 7.5 minutes
Big plane example:
8000 mAh battery
power system draws 64A
8000 mAh = 8.0 Ah
8.0/64 x 60 = 7.5 minutes
As you can see, the small plane drawing 12A will need a 1500 mAh battery to fly for 7.5 minutes. The big plane drawing 64A will need a 8000 mAh battery to fly for 7.5 minutes.
- Jeff
