SPFX Camera Dolly - Motor Acceleration Amps 1

[eckener]

Hi All... We are building a special effects film dolly. The motors have been sized and we are trying to estimate the current draw during the very fast accelerations... Doesn't seem to be too problematic for constant speed segments of our move profiles, but figuring out the current draw for accelerations has us a bit stumped.

Is there a relatively easy short-hand way to do this? perhaps using the torque constant (Kt)of the motor? or another formula that we have just been missing.

We have already figured out all our mechanical loads, wheel frictions, air drag forces, moments, gear- ratio, etc... We have this all in a spreadsheet where we plug in the acceleration rate and top speed. For simplicities sake, it spits out us a constant torque value on the motor shaft over that period of time to achieve that desired acceleration...

Now we want to find out how many amp hours that will approximately take.

We tried a couple motor sizing softwares, but they really didn't give us this information.

We did convert the whole shebang (including losses) into Work Done(Joules), then into Watts and then into Amps. When we compared that to the simple equation, Torque/Kt, the Torque/Kt was about 2-1/2 times higher... maybe this is because the voltage was not changing in this equation? Should we cut the voltage in half?

Its a brushless DC Servo motor (actually 4 of them.. Kollmorgen AKM series)
320Volt DC Battery Pack
Elmo Drivers
200 pound dolly with payload
3.91:1 gear reduction on motors.
6" diameter drivewheels.
27.4.mph top speed
14.65 ft/sec/sec accelerations (typical)

Thank you for reading.

 

[waross]

At a given speed the back EMF will be close to the applied voltage. The difference between the applied voltage and the Back EMF results in current through the motor and the source. Generally the source and the armature circuit resistances are very low.
An example:
Ignore the source resistance for now.
Assume that the applied voltage is 12 Volts and the back EMF is 11 Volts. The armature circuit resistance is 1 Ohm. One Volt will drive one Amp through one Ohm.
Now double the applied voltage to 24 volts. The difference between the back EMF and the applied voltage is now 24V - 11V or 13 Volts.
You may expect the initial current to be 13 Amps or 1300%.
As the motor accelerates the back EMF will rise and the current will drop.
However, at 1300% Current the source internal resistance may reduce the terminal voltage somewhat.
This word description may help to understand and estimate the voltage-current relationships in an accelerating DC motor.
Do not use my value of the armature circuit for estimates. Actual values will vary widely from motor to motor.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[eckener]

That's a great description! and very helpful... really helped me understand. Thank you.

It seems to explain the current as a "snapshot" within the acceleration however. Perhaps to get a total amp-hours for the entire acceleration, I could calculate with this method at, say, 10 intervals of the voltage rise through an acceleration, then calculate each of those currents in terms of the total voltage of the battery pack.... then sum them all together.

I was initially trying to approach a different way I guess, by calculating the work done by the acceleration, then applying a motor/driver "efficiency" to it.

 

[iop995]

If you have torque (M) value and time (t), to compute energy. If acceleration (a) is constant, motor acc will be eps=a*k/60/d, where k is gear reduction and d is diameter drivewheels; max motor speed is Omega_max=eps*t, so enegy developed is W=M*Omega_max/2; correct with motor and converter efficiency and obtatain energy from source. Depending of source type may compute current (average or maxim) nedeed.

 

[eckener]

hmmm... that one I couldn't follow. sorry, but thanks.

 

[SceneryDriver]

Perhaps I'm incorrect, and this would be very conservative, but locked rotor amps (or equivalent, amperage at 100% torque) for the duration of acceleration could be a value to use. Like I said, LRA/100% torque would be overkill, but it I think it would give you a worst-case scenario.

What amperage can the servo controllers handle without tripping on over-current? That may be a limiting factor too.



SceneryDriver

 

[waross]

Remember, as the motor is accelerating, the back EMF is increasing and the current is dropping.
If you are using a servo controller it will probably be in current limit for hard acceleration.
You need some information about the controller.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[tygerdawg]

And here's a proposal from a dumb Mechanical Guy who has done quite a lot of precision motion control over the years and who isn't smart enough to get into the nuances of motor Electrical Engineering (them electrons scare me):

I suggest you should perform your analysis based on the maximum torque required to move your loads. This involves linear accelerations/decelerations of linear masses, rotational accels/decels of rotary masses, accounting for gravity effects, friction/stiction, dirt on the bearing, torques induced by offset loads, all kinds of things. Develop a value for "PeakTorque". Size your motor for PeakTorque*2 or 2.5 and run with it. Excess torque is not necessarily a bad thing, it gives you strength to overcome loads you never anticipated. It's there in reserve if you ever need it. Make sure your mechanical transmission elements won't get damaged if max torque is applied.

Current draw is amp load at max torque (which you should never achieve anyway).

Just a suggestion.

TygerDawg
Blue Technik LLC
Virtuoso Robotics Engineering

http://www.bluetechnik.com

 

[iop995]

To move a load at constant acceleration need a constant torque all time (considering load = constant vs speed). For a DC motor that mean constant current (considering constant flux), so as speed increase, to keep constant current (torque) need that converter increase applied voltage; current being constatnt will impose a constant acceleration and power develoved by motor increase liniary up to maxim speed. Sure all mechanical parts and power converter need to be designed accordingly requested acceleration and load values. If source energy have constant volage (battery), current suplied will have same shape like power, will increase from a start-up value up to maxim corresponding with point very before acceleration become zero (max speed).

 

[eckener]

Lots of info here... but ughh! Honestly I've tried for 4 days straight: I've looked at literally hundreds of white papers and they are either way over my head technicaly or seemingly have completely different equations.... either way none of my attempts to calculate the amp draw are coming in remotely close to reality, and Kollmorgen sizing software (motioneering) has nothing about the amperage. (and I've been very careful on my units.) The motor is sized correctly for the application in terms of rms torque, and peak torq..


Maybe I will simplify my query, lets say we forget about acceleration, and just ask how to somewhat accurately calculate the current draw for a steady state mode....

Here's my example...

motor speed: 6000 RPM (continuous: no acceleration)
torque on motor shaft: 3.721 Nm (continuous)
WHAT IS THE CURRENT DRAWN FROM THE BATTERY?

In a purely theoretecial calculation of the mechanism (but complete with friction losses), it would take take from the motor 2598 Watts of power, 8.12 Amps.

But of course the motor and driver are not 100% efficient.

Battery voltage: 320VDC
Motor Torque Constant: .43 N-m/Arms
Motor Back EMF Constant: 27.5 V/krpm
Motor Resistance (line to line): .8 ohm
Motor Inductance: 3.1Mh
Motor Continuous Torque Stall at 100 deg: 3.53 Nm
Motor Continuous Current Stall: 8.4Ics
Motor Max Mechanical Speed: 6000rpm
Peak Torque: 11.6 Nm

 

[eckener]

Don't know if this matters, but the motor has 5 pole pairs.

 

[Sparweb]

...Kollmorgen sizing software (motioneering) has nothing about the amperage.

Well maybe the sizing software package doesn't try to tell you this, so instead look through the controller's datasheet. It really is important for all designers to know the maximum current that can be driven by a controller, and the maximum current draw in puts on the line. Wire sizing, fuzing, circuit breakers, insulation, all that stuff. I think the point you're missing, which a number of members have tried to say already, it that the maximum current your system will see is the maximum it can handle before letting out the "magic smoke" or blowing a fuse. This information is so important to electrical design that I'm surprised you haven't just read it off the datasheet and moved on by now.

Very first hit on google, using "stepper motor 320VDC" is a datasheet with torque-current plots and a max current statement all on one page.
If you are NOT seeing this from the devices you have selected, then you have more questions to ask the manufacturer.

I tried looking for a Kollmorgen servo, took a minute longer, found the catalog for one of their product lines, and just poked around. I picked the AKM but I don't know what you are using. About 1/3 of the way through the nicely detailed specifications, I found current at stalled torque under numerous conditions, and a peak current with its corresponding torque... It looks like the number you want. I won't deny that it can be time-consuming to sift through the mountains of information these catalogs can pile on you - this one is 70 pages long.



STF

 

[eckener]

Sparweb, I'm not interested in the "maximum the controller can handle" and have never mentioned that. I know we are under that limit. Rather I am interested in calculating amperages of various lesser motor accelerations so that we can size the amp hour capacity of our batteries according to various situations.

And yes, I've seen the curves: our motor is sized correctly: what I wanted was an underlying formula.

I do believe that iop995 answered the question the best, basically informing me that a current curve would basically mimic the power curves no matter what... (he didn't say it would mimic the torque curve, but I think that might be what was implied)

But nobody here has mentioned the following formula:

Amperage = Torque / K_t (torque constant.)​

But I think maybe its just that simple.
And If it is not, please feel free to correct me...

So therefore, If I am simulating a very simplified acceleration with a constant torque value, then, as iop995 says, the corresponding current would also be continuous single value. And if this number was divided by the Kt of the motor:

I=T_l/K_t

... and multiplied by the time of the acceleration and then divided by the battery voltage,

A_h=I*s*/V

where:
Ah = the amp hours used during the acceleration
I = the current
s = the time in seconds for the acceleration
V = the battery voltage
I is the amperage
T_l is the torque required to achieve the acceleration
K_t is the torque constant of the motor

I think this is right... 90% sure... If anybody thinks diffently, please do set me straight.

What really threw me originally is that when you calculate the energy, or Work done, of the mechanism undergoing the acceleration, then convert it to watts and then amps, you guy a pretty low number. I thought that the motors, being fairly efficient would only be a little bit above this, but doing what iop995 suggested is considerably higher... 2.4 times higher in my case, and that really through me off.

If these motors are suppposed to be more than 80% efficient, then why is it only like 40% efficient in this acceleration?

 

[Sparweb]

You sound frustrated, and now that you've explained some more, I see that you are past the selection and sizing of components - a process I thought you were still undergoing.
I hope to understand why you are going through this analytical exercise.

Going back to review your earlier postings, I think I can glean some performance requirements from the mention of amp-hour capacity and the battery packs....

Would this be a correct statement, on your part: "I am trying to determine the battery capacity required to satisfactorily power this system at a remote location, without chance of recharge or resupply, for the period of time usually required for a film crew to complete their task, under all foreseeable conditions". Two corollaries of this are that too little battery capacity means the shoot is cut short, and too much battery capacity means carrying dead weight to the four corners of the earth. If this is the case, then I understand your goal much better, and we can dispense with the explanations of wire sizes and over-current devices - which are the usual subject of discussion when "maximum current" is called into question.

Have you tested the system to take measurements yet?


STF

 

[iop995]

eckener: Formula Amperage = Torque / Kt (torque constant.) tell that motor current is constant for a constant torque, but you need to calculate (estimate) Ah battery; as motor speed increase, power developed increase also. At converter input, voltage battery being nearly constant, current supplied by battery will increase also.
For a given torque, at start-up speed (maybe zero) you have motor current I, compute power developed by motor, correct it with motor and converter efficiency and obtain battery requested power (Pbat) and battery current (Ibmin=Pbat/Vbat), then this current increase liniay with speed and have Ibmax at maxim speed. Or, average current will be Ibavg = (Ibmax + Ibmin)/MaxSpeed/2.
To calculate Ah battery, in this case, when current is not constant, I thing best way is to check battery discharge time at constant power and using a 65-70% from maxim calculate power will lead to a good estimation (or may use maxim power to have a reserve room).

 

[eckener]

hahaha... thanks SparWeb.. can you tell? Most of the systems on this dolly, we have a good handle on, but one of the dangerous of the entertainment industry, and especially of VFX and SPFX supervisors is that we are jacks of all trades, masters of hopefully at least few. I've a solid handle on most of the engineering aspects of this project, but calculating motor at this level of detail is not something I've done before.

And yes, that would be an accurate description of at least one of the reasons we are trying to determine amp-hours carefully! Other performance feats we are calculating include for instance, motoring up and down a football field at 27 mph at 14 ft/sec/sec, and making sure we have enough battery life to get to halftime, or calculating amps required to go straight up the side of a building for the next Spiderman movie.. or shooting a car stunt at zero to 60mph in less than three seconds...

As this is a semi-autonomous vehicle, its a delicate balance of weight, bulk and performance and above all, safety. Batteries are probably one of the costliest and bulkiest and most critical components to the system. (We are currently using a 2.4 amphour 320V LiFEPO4 lithium pack)

But another reason we are trying to nail down the amps is to make sure our drivers can do the job, but are not so big that they increase the volume too much.

We are just embarking on building on prototype now, but haven't ordered parts yet. Wanted to try and get the performance envelope and physical envelope nailed down as much as possible before ordering...

After figuring out the amp usage here, the next thing we are tackling is how to best stop the thing. (handle our back-EMF): deciding whether to use quite bulky and very hot dynamic braking resistors, or whether its worth it to regen into the battery.

 

[ScottyUK]

As battery life is important, and given that this thing will have a fair amount of kinetic energy, then a regenerative system must make sense. In terms of additional mass it won't add much to the driver, and probably less than a braking resistor bank.

 

[iop995]

With a lot of simplifications, attached file try to made an estimation. I hope there are no errors, please check.

 

[eckener]

Iop995, Thank you so much for your help. Please forgive me for rewriting your sentences a little bit…

The formula I= T / Kt tells us that motor current is constant for a constant torque, but you need to find the amp hours for the battery:

As motor speed increase, the power developed also increases. At the converter input, voltage battery being nearly constant, the current supplied by battery will increase also.

Ok… so if I am getting this right, you are saying that even though the current and torque is remaining the same, it’s the watts of power being generated by the motor over the range of RPM that will affect the Ah calculation…. and I guess this assumes that at different RPMs the motor will be using less than the full voltage of the battery. (but I am not sure about this, because of the way the drivers control both voltage and current.)

However, I think the way I am doing it now, I'm assuming a max of 320 volts over the whole acceleration, (which would be maximum power over the whole acceleration.)

The method you descibe I think would result in less Ah than the way I am calculating it now…


For a given torque, at start-up speed (maybe zero) you have motor current I,
compute power developed by motor,

If this same as calculating the power required by the mechanism over time, I had been doing this by calculating the mechanisms energy in Joules divided by the duration of the acceleration in seconds, however this gives an average power. Or are you saying the power is based on the current inside the motor at a certain torque?

correct it with motor and converter efficiency

I was not finding this efficiency rating specifically listed in any documentation, and was trying to use just multiplying the value of K_t, but this is probably not correct?

and obtain battery requested power (Pbat)

This is where I was really having difficulty. To find watts here, based on torque and rpm, don’t I need the voltage that the motor is using at this rpm…. How to find it? I tried using K_v, but it seemed to be giving me only half the number of volts at full RPM of the motor

and battery current (Ibmin=Pbat/Vbat),

then this current increase linearly with speed and have Ibmax at maxim speed.

Or, average current will be Ibavg = (Ibmax + Ibmin)/MaxSpeed/2.

To calculate the Ah from battery, in this case, when current is not constant, I think the best way is to check battery discharge time at constant power and using a 65-70% from maximally calculated power will lead to a good estimation (or may use maximum power to have a reserve room).

I believe the method I am using now is the maximum power method, as I am assuming that the motor is always sucking up 320. But would like to be doing this the correct way.

 

[eckener]

iop995.. you are awesome! I will check out that spread sheet later on today
Thanks!