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Systema broke down on second mag!

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Old October 14th, 2007, 15:35   #16
MadMax
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The PTW has a brushed permanent magnet DC motor. You can do funny stuff with the electronics, but while it's driving forwards, it doesn't make sense to modulate voltage.

I'm not sure how you come to the conclusion that the PTW is a constant power device. Such an assumption would imply that a lower voltage battery would be able to drive the same ROF with equivalent spring. I find no buck boosting circuitry in the switch FET assembly to step up the output voltage and make the PTW constant power output regardless of input voltage.

A lower voltage would result in a lower rotation rate at similar torque requirements.

http://www.joliettech.com/dc_motor_c...cteristics.htm

Current would be determined by:

I = (input voltage - back EMF)/(coil + brush resistance)

Output torque = Kt * I [motor torque const * I]

With a non changing spring set, the average torque requirement would not change. This results in a similar current requirement for different voltages.

The main spring does not approximate a reactance very well. Energy stored in it is not returned to the system as it would in an inductor. It's instantaneous impedance charactaristic is also not like a typical coil reactance as it does not offer a impedance proportional to the rate of compression.

In any case, an AEG motor firing full auto does not experience very high varience in load torque while it spins through the spring compression cycle. The gear reduction allows the motor to spin a pretty high rotation rate which stores a lot of inertia in the armature. The armature acts as a flywheel which smooths out the rotation rate significantly which results in a surprisingly more or less constant torque operation.

This was discovered in the exploration of AEG control circuits on Airsoft Mechanics:

http://forums.airsoftmechanics.com/i...p?topic=648.15
http://team-titanium.com/~airsoft/Co...5/tek00000.gif

Over multiple shot cycles, the input current is pretty consistent. It was initially hoped that monitoring motor current could be used to determine sector gear position to make a solid state shot counter (burst feature!) with no direct sector gear observation. However, what is detectable is the current fluctuation during commutator pole change. It might be possible to monitor the motor comm to infer the sector position, but I have a feeling that reverse rotation before the anti reversal latch catches may screw up the "accounting" of comm poles.

One other thing occurs to me: the higher rotation rate pushed by the 12v battery results in more inertia storage in the armature flywheel. Increasing input voltage by 25% (12v vs 9.6v) would increase RPM by 25%. A 25% increase in RPM would store 56% more rotational energy in the armature (energy related to the square of rpm). This would provide a bigger repository of inertia to power through the spring cycle.

It still bothers me that SystemA is shipping stock springs that are so close to the limits of operation. A product that pops it's fuse on a dead battery isn't very robust. Perhaps more like a finely tuned finicky race car than a tank. Robust electrical design would have the fuse rated to blow at about 85% of the stall current draw of the motor. Instead it protects under spec'd electronics that may not be able to power the motor to 85% of stall torque. Peak power output on a motor occurs at 50% of stall torque so it's important to design a motors electricals to handle over 100% of stall torque (current) and fuse at around 85% to protect the motor from overcurrent at stall.
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Old October 14th, 2007, 17:02   #17
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Quote:
Originally Posted by MadMax View Post
I'm not sure how you come to the conclusion that the PTW is a constant power device.
I neglected to include that:

Current (I) = Voltage (V) / Resistance (R)

Power (P) = Voltage (V) squared / Resistance (R)

As impedance (Z) equals resistance (R) in a DC system, and you do not change the spring load, dropping voltage means increasing current. Motor torque equations don't seem to apply here. I do believe these motors to be of a constant power design, as are most motors in the AC world.

I agree that Systema could design more robust electronics, but as far a fuse ratings go, the CEC requires that the maximum fuse size for a one time fuse is 300% of FLA (full load amps) and that motors must have an FLA rating at 1/6 of LRA (locked rotor amps). Time delay fuses cannot be at more than 175% of FLA. Now, the CEC applies to motors for industrial, commercial and consumer applications and only in the jurisdiction of the CSA. Systema Japan does not have to bow to our electrical system.

I would suspect that Systema has their rating at about 125% of FLA, but to design it to be more robust would mean a switch board that would be 2 or 3 times it's current size and not practical for a collapsible stock.

The spring does not approximate reactance, the motor does this at each pole reversal. What begins as a DC impluse at one pole, then short rotation as rotor turns to attracted fixed magnet pole, then pole reversal and simultaneous repulsion from current pole to attraction to next pole. It makes the DC current and voltage look very much like a sine wave (AC). This AC is what turns an inductor into a reactor, making change in reactive current difficult (an inductor wants to limit changes in current). In this case, it simply can't as motor windings heat up (alot on the M150 for those unfamiliar with the PTW), and this causes current to rise even more. It's a vicious negative cycle that only increases the more you fire.

Perhaps I typed slower than I was thinking. I fxed my previous post. The spring could be thought of more as a resistive load, seeing as how it likely stays uniform through most of it's compression. But that is neither here nor there if you are retaining the same spring but varying voltage, either by changing from a 12 volt to a 9.6 volt battery, or by have your battery become depleted. My analysis is supported by the fact that the fuse will blow, or the switch will fail. These are certain and documented events that have occurred in the past and will occur in the future if people fail to heed the warnings of a low battery or use the incorrect voltage/cylinder match.
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Last edited by mcguyver; October 14th, 2007 at 19:14..
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Old October 14th, 2007, 21:05   #18
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Originally Posted by KEVORKIAN View Post
I chose to pick up my PTW Max CQB from MadMax & I have yet to have any problems...The money I saved by having it QC'd & built by him is well worth the added cost! I even got the 6-pack of mags in the deal !!! Quit wasting your money TRYING to save money...

Let this be a lesson to all who are looking into purchasing a Max SCK...
yeah a teammate did the same and its been flawless.
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Old October 14th, 2007, 21:34   #19
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Thanks Madmax for posting because you have solved my problem. Turns out it was the adjustment screw at the bottom of the grip, with the motor running the screw turns! In the instructions it says to put the screw in but to only put it flush with the motor plate, however even with the screw above the motor plate it still turns. I dont know why Im the only one who would of had this problem. Anyways all is well thanks again!
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Old October 15th, 2007, 00:02   #20
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Quote:
Originally Posted by pawscal View Post
Thanks Madmax for posting because you have solved my problem. Turns out it was the adjustment screw at the bottom of the grip, with the motor running the screw turns! In the instructions it says to put the screw in but to only put it flush with the motor plate, however even with the screw above the motor plate it still turns. I dont know why Im the only one who would of had this problem. Anyways all is well thanks again!
So you found that the screw had tightened itself after wandering forward and touching the motor shaft?! HAHAahhaa. I have to admit it was a wild ass guess. Back the screw out and loctite it so it stays clear of the motor shaft.

Usually something has to change for something to stop working. I had to think about the series of stuff in a PTW to come to a few places which could change and cause an increased electrical or electro mechanical load.

Quote:
I neglected to include that:

Current (I) = Voltage (V) / Resistance (R)

Power (P) = Voltage (V) squared / Resistance (R)

As impedance (Z) equals resistance (R) in a DC system, and you do not change the spring load, dropping voltage means increasing current. Motor torque equations don't seem to apply here. I do believe these motors to be of a constant power design, as are most motors in the AC world.
You're quoting Ohm's law which is incomplete for the analysis of DC permanent magnet motors. Ohm's law is useful for calculating current in a stalled motor, but once a motor is spinning, an effect called the Lorentz effect generates a back EMF voltage potential which starts to counter the input voltage to the motor.

Back EMF in a motor is quite literally the voltage generated by the rotating armature. If one spins a motor shaft with a drill, one develops a voltage at the moter terminal. If you apply a voltage to a motors terminals, the armature speeds up until the back EMF developed by the armature balences the input voltage.

In this condition (on an unloaded motor) current flow is very low (some energy draw to beat windage, resistance and brush losses, but still very low) because the back EMF component nearly equals the input voltage so the only current drawn is due to the imbalence in back EMF divided by coil resistance. Hence:

I = (Vapplied - EMF)/ (coil resistance + brush resistance)

EMF = motor constant * rotation rate in radians

This phenomenon of back EMF is the reason that motors don't continuously acclerate until they exceed their winding strength and explode due to centrifugal acceleration when no load is applied to the shaft. At some point the back EMF balences the applied voltage so current draw drops and little power is drawn.

http://lancet.mit.edu/motors/motors3.html#speed

The last chart shows the behavior of a DC motor under constant voltage condition with varying torque output.

Under no load condition, the motor rotates at 600rpm and exerts no torque so it outputs no power. Under stall condition, the motor rotates at zero rpm and exerts 0.3Nm torque and still outputs no power (force but no revs). Maximum power is output at half stall torque and half no load rpm. Clearly DC perm magnet motors are not constant power output devices. Their load condition significantly impacts their power output and associated power draw.

Quote:
I would suspect that Systema has their rating at about 125% of FLA, but to design it to be more robust would mean a switch board that would be 2 or 3 times it's current size and not practical for a collapsible stock.
I don't know about that. There are MOSFETs in the same packaging as the ones that SystemA uses that are rated to 72A. With four in parallal, they'd be good to 288A which is a huge load higher than what the 30A fuse handles. I suspect that the FETs are not dying because of current overload. I have a feeling that they're getting killed by an inductance EMF when the motors are actively braked. The inductance EMF generated when the field collapses around an armature coil can be quite high. R/C car speed controllers carefully snub motor back EMF when braking to protect the PWM circuitry. I have a feeling that SystemA hasn't done a very good job with this. R/C cars and airplanes put some very high electrical demands on their power electronics. SystemA should take some controllers apart and see what other Japanese hobbyists are putting in their products.
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Old October 15th, 2007, 00:22   #21
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Yes, Max, I know there is more to the analysis of motor dynamics than just Ohm's Law, but when you factor in cemf being present in a running motor (which is what we're talking about here), all you are left with is simply voltage, current and resistance. The reactance will stay constant, unless you vary frequency of pole reversal. So that being said, Ohm's Law is most applicable to that actual fact of blowing fuses (overcurrent) when the applied voltage is too low for the load rating. When I have to design or implement a motor control system, I don't factor cemf into the equation because it doesn't matter while the motor is running. And cemf will never exceed inrush (I-squared x R losses). Usually, when I have to do these calculations and work, it can range from 10 amps to thousand of amps at voltages up to 600 Volts. Luckily, I have charts to do all the hard work for me.

And, at 125%, I mean fuse rating above FLA of the motor.

I agree, that cemf presents a problem with braking, and perhaps the FETs that have failed for others have not solely been the fault of the FET itself, but the design of the braking system, even the quality of the circuit board itself. But low voltage is still a problem for this system, hence the recommendations from System on battery/cylinder matching and their increased FET capacity from the Gen 1 100 amp and 7.2 volt system to todays MAX at 12 volts and 200+ amps in FET capacity.
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Old October 15th, 2007, 01:57   #22
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Quote:
Originally Posted by mcguyver View Post
I neglected to include that:

Current (I) = Voltage (V) / Resistance (R)

Power (P) = Voltage (V) squared / Resistance (R)

As impedance (Z) equals resistance (R) in a DC system, and you do not change the spring load, dropping voltage means increasing current. Motor torque equations don't seem to apply here. I do believe these motors to be of a constant power design, as are most motors in the AC world.
I'm still stuck on this bit.

How are PTW motors constant power? Running at a lower ROF with a lower voltage motor clearly outputs lower power as you get fewer equivalent energy shots per minute and therefore lower average watts.

If this is the case, why would current have to increase to match to drop in resistance? If a motor takes more time to complete a firing cycle, it seems pretty apparent to me that it's running a lower power output to accumulate the same work in the spring.
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Old October 15th, 2007, 02:08   #23
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Originally Posted by MadMax View Post
I'm still stuck on this bit.

How are PTW motors constant power? Running at a lower ROF with a lower voltage motor clearly outputs lower power as you get fewer equivalent energy shots per minute and therefore lower average watts.

If this is the case, why would current have to increase to match to drop in resistance? If a motor takes more time to complete a firing cycle, it seems pretty apparent to me that it's running a lower power output to accumulate the same work in the spring.

This is the only explanation I can think of why lower voltage input means more current input. You see these things occuring in other areas as well with motors, but usually it has to do with VFDs driving motors or another example that comes to mind is the power supply for car audio amplifiers.

By your train of thought, lower voltage should not matter, but the empirical evidence of fried electronics due to low voltage (barring any other variables that nobody has cared to mention) seems to be conclusive, in my opinion.

I don't know. I see what I see and apply what I've learned on the job and in school. What more can I say?
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Old October 15th, 2007, 02:47   #24
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Well, I didn't say that lower voltage didn't matter. I meant to say that it might matter in other ways than constant power driving a higher current. Clearly if there are repeatable failures with low voltage packs, then there's a definite overcurrent condition. I'm just exploring the possible causes.

Almost all of my design experience is with low power constant DC motors. I've hooked up some sub hp VFDs and played with shaded pole and other motors, but I've not dived into their subtleties. School covered them, but I haven't had to put them to enough use that the lessons really sunk in.

I had a tough time figuring out how lower input voltage might result in higher current given the prevalent constant current-torque behavior of DC motors. I did think of pole switching, but the specs on fry time on the fuses was too long to respond to the fast current changes in pole changing. I could only guess that there was a long one pellet shot cycle issue at hand. An issue that was longer than the two dozen or so revs on the motor required to crank the sector thru the spring cycle.

I'm starting to lean towards the lower voltage = lower rotational energy stored in the armature guess as a faster armature might smooth out current draw over the spring cranking cycle a lot.

Come to think of it, if the motor slows down significantly during the crank cycle, the CEMF drops proportionally which would quickly ramp up current draw. A motor that is initially spun up fast by a higher input voltage (in the easy low compression windup) would slow down proportionately less. Current draw at the beginning of the start up would be higher, but it could be lower when cranking the spring because the power invested in the early part of the cycle would help push past the spring.
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