How to calculate amps drawn?

[EstebanUno]

I've been poking around the board for a week or so soaking up information, and I remain unclear about how to calculate how many amps are drawn from a battery at different loads. Hence my first post.

I'm trying to calculate how many amps are drawn when a motor is run at an efficient rpm with varying loads. Assuming the motor is running at an efficient 80% speed, the load can vary based on grade, wind, pedal assist, etc. For example if the output power required to maintain a speed of 25mph is 400 watts (let's just assume for the example that 25mph yields an 80% efficient rpm), and the system is 36v a simple calculation yields 400/.8/36=14 amps. Roughly speaking, is it as simple as this? Can I just plug in the output required to estimated amps drawn, and then judge how many amp hours my battery should yield?

If so then with a big tailwind, the output required could be halved to maintain the same 25mph speed, meaning only 7 amps would be drawn. Or conversely, 28 amps might be needed with a strong headwind or incline. If so, this implies a linear relationship between load required and current drawn, and I'm wondering if that is the case?

Esteban

 

[katou]

It's not that simple, although we all wish it was.

In theory, everything theoretical works.
In practice, it practically never does.

The motor efficiency varies with RPM, and the amount of amperage available from the battery is not infinite.

So, the motor is not necessarily getting what theory says it should from the battery
You don't know the RPM at every second.
How much the motor actually draws from moment to moment is also unknown.

Basically, you just have to chuck the calculator (except as a rough guideline) and measure the info that you want.

Most use a Cycle Analyst bike computer to do that. It's pretty much the only way that I know of that's easy.

Katou

 

[Miles]

The motor efficiency varies with RPM,

Actually, it's a bit misleading to say this. Strictly, it varies with torque/current. Peak efficiency is achieved when the copper losses equal the parasitic losses. The motor's torque constant will give you the linear relationship between torque and current. During PWM it's the phase current that counts.

 

[EstebanUno]

It's not that simple, although we all wish it was.

In theory, everything theoretical works.
In practice, it practically never does.

Understood, but is the theory as I presented it correct? To use a rough guideline, assuming steady rpm and adequate current from the battery.

There is a 4 mile 2000 ft climb that would be part of my commute. I'm considering a Cyclone system, and need some reference for battery draw. I've calculated power needed to to make the climb at 10mph at the nifty web site http://www.analyticcycling.com/ForcesPower_Page.html, I came up with 436 watts. I'm trying to get an idea what battery would be required to survive 25 minutes of almost continuous draw at 10 mph, or 15 minutes at 15 mph. I've got no real-world reference other than my own human power can climb it at 6-7 mph, which calculates to 250 watts and leaves me exhausted.

Esteban

 

[katou]

As usual, Miles is technically correct. After a year hanging around here, I have very nearly the ability to understand what he's talking about.

I understand that there is a relationship between kv, kt, torque, amperage, and I even know sort of, what the relationships are. The puzzle piece I'm missing is WHY the relationships exist, why they are connected.

In any case, your problem is the slope entirely. Most hub motors would not survive that grade.

I would think that your question should best be put to the makers of the mighty Cyclone system. They should be able to tell you if the motor and controller has a duty cycle that will survive your grade.

I am guessing that you want to know about the battery draw to determine what size of battery to purchase?

Other factors may determine battery for you. You may not be happy with the performance around town with a 2c rated (Pings I understand are 2c rated) Ping pack. You may not have the money for a Ping pack. You may not have the time/interest to put a pack together out of cells. You may not like the weight of the Ping and opt for a lighter, lithium polymer pack. These may be bigger decisions than just amps drawn.

My point is, the motor may not survive, and it's hard to know without trying, so contact the man. and find out if they will honor warrantee if it dies in your use.

HTH,

Katou

 

[Miles]

I understand that there is a relationship between kv, kt, torque, amperage, and I even know sort of, what the relationships are. The puzzle piece I'm missing is WHY the relationships exist, why they are connected.

Reading this was my Aha! moment: http://groups.yahoo.com/group/lrk-torquemax/message/7728


Let
Kt = torque constant in Nm/A
Ke = generator constant in V/(rad/sec)
V = actual winding voltage (= Vsupply-IR)
I = winding current
w = shaft speed in radians/sec
T = generated torque (available torque is less, of course)

From conservation of energy, electrical input power to windings must
equal the mechanical power produced by the windings. That is :-

VI = Tw
thus V/w = T/I
ie Ke = Kt

It has to be so, or energy would not be conserved.

 

[dogman dan]

You can just go out and find out. Ampmeters can be as cheap as a car type ampmeter from harbor freight, 5 bucks.

I find you can do some generalization. Your figure for 25 mph is a bit off I think, about 500-600w is more like it. 350- 400w gets you 20 mph. 15 mph only takes about 200-250w. These figures seem to be about the same for many types of brushless motors. Brushed motors use more, and above 25mph then things get affected enough by wind resistance to get more vague. This is what I get on bikes with no particular aero advantages, IE, sitting up tall on an MTB or beach cruiser.

For headwinds or tailwinds, the numbers change accordingly, 20 mph with a 5 mph tailwind rides like 15 mph power use. And so on.

This is useless for those who want really accurate calculations, but in my world, that kind of accuracy doesn't exist. That's exactly why you oversize your battery. With about 25% exess size, I get home no matter how much headwind I have to ride uphill into.

 

[neptronix]

Find out with an ammeter </thread>

turnigy watt-meter is cheap and tells you the voltage and AH used also.. quite nice.

 

[TylerDurden]

The topic title is a bit misleading, it might be helpful to revise... "help estimating battery capacity" or such.

Your projection gives a rough estimate.

Efficiency will likely be lower due to motor heating, voltage drop and other losses. (Maybe 50%.)

That's about a 10% slope, requiring >550W @ 10mph at the wheel.

1100W for .4hr from the batteries = 440Whr, before de-rating for Peukert and DoD.

That's my "guesstimate" FWIW. There are some roadies here that might have some bona-fide data. (JennyB?)

 

[katou]

And then there's a very sensible method to estimate, we'll call it, "buy what you can and make it work" method.

Buy a battery that you think is pretty much going to make it, then just pedal as much as necessary to leave you enough juice to get home.

If I was in doubt of my battery, I'd just pedal as necessary to keep the system happy. Just because I CAN ride without pedaling, doesn't mean I have to.

Thanks for the info Miles, I will look at that very closely and try to understand. I haven't checked out the link yet. Do you know of any worked examples?

Katou

 

[EstebanUno]

The topic title is a bit misleading, it might be helpful to revise... "help estimating battery capacity" or such.

Your projection gives a rough estimate.

I think the thread topic expresses what I want to know, though I do appreciate the real-world estimates. I'm a complete novice when it comes to electricity. The basic concept I was after was whether the power required mathematically determines the current in amps drawn, if rpms are fixed. And whether it is a linear relationship as the power requirement varies. Perhaps the following question phrases the concept more succinctly: Does the motor alone at an efficient rpm draw twice the amps than it would if pedal assisted for half the power requirement? I'm still not clear about that.

Esteban

 

[TylerDurden]

Does the motor alone at an efficient rpm draw twice the amps than it would if pedal assisted for half the power requirement?

Yes (approximately).

Power = Voltage x Current
(Watts = Volts x Amps)

 

[EstebanUno]

[/quote]
Yes (approximately).

Power = Voltage x Current
(Watts = Volts x Amps)[/quote]

Ah, yes. Thanks for the insight.

 

[katou]

Ah, I see, your question was about the theory of it all.

Here's an example:

If the job requires 500 watts, and you contribute 200, the remainder left for the motor is 300.

Power (watts) = V x A

work = power * time

The real tough part here is that you must calculate based on work, and it is difficult to estimate how much work the human element is performing.

It would be much easier to calculate based on rpm, because it is visible and easily measured, but that is only part of the equation:


Power (kw) = [torque (nm) x 2 x Pi x rpm] / 60,000

And knowing the torque of a pedaling person is very, very hard to measure without some fancy equipment.

HTH,

Katou

 

[EstebanUno]

Ah, I see, your question was about the theory of it all.

Here's an example:

If the job requires 500 watts, and you contribute 200, the remainder left for the motor is 300.

Power (watts) = V x A

work = power * time

The real tough part here is that you must calculate based on work, and it is difficult to estimate how much work the human element is performing.

It would be much easier to calculate based on rpm, because it is visible and easily measured, but that is only part of the equation:


Power (kw) = [torque (nm) x 2 x Pi x rpm] / 60,000

And knowing the torque of a pedaling person is very, very hard to measure without some fancy equipment.

The way I tackled the human element is through the power-speed-force calculation that I linked earlier in the thread. Since I know my weight, my bike's weight, the incline and the speed I can attain at various exertion levels, I can find the theoretical power to the crank I'm providing with the calculator, in watts. One can use mapMyRide to get the grade of a long, even grade and watch the cyclo computer for speed attained at various exertion levels.

I used the same calculator to estimate power in watts required for speeds I cannot attain that should be possible with the motor, or motor + me. According to the feedback on this thread, the calculation is on the low side. I'll keep that in mind and pad the values I get.

Esteban

 

[TylerDurden]

Here is the chart I have used (probably for too long). I used data from an online calculator.
Rough, but useful.