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Old October 17th, 2009, 03:14   #61
Kos-Mos
 
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Quote:
Originally Posted by iggim109 View Post
ok so with this being said i was wondering that is this: http://shop.ehobbyasia.com/bb-batter...ak-series.html
worth it to buy???
You know the answer. You QUOTED it.

There is over 70 types of lithium batteries. Most are NOT suitable for high discharge use (think watch battery or photo camera battery).

Lithium =/= awesome.

Now stop polluting this thread with questions that HAVE BEEN ANSWERED ALREADY IN THE SAID THREAD!
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Old October 29th, 2009, 19:23   #62
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Updated with more information about LiPos and why use a properly rated battery.
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Old October 29th, 2009, 21:56   #63
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Thank you...that's awesome.
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Old October 29th, 2009, 22:05   #64
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Ebay chargers are good for your money. I have 2 of them and i can tell you, they are worth your 60$ (they can charge ANYTHING)

If you want to use a LiPo but your gun is stock, best is to use a battery which gives discharge rate of 15C and at 7.4V. 7.4V in lipo is equivalent to 9.6V nimh ... 15C will not tear your gearbox and gears apart like an 11.1V , that's why most often, 7.4V 15C batteries priced like an 11.1V 20-35C battery.
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Old October 29th, 2009, 23:15   #65
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Originally Posted by Ayashifx55 View Post
Ebay chargers are good for your money. I have 2 of them and i can tell you, they are worth your 60$ (they can charge ANYTHING)

If you want to use a LiPo but your gun is stock, best is to use a battery which gives discharge rate of 15C and at 7.4V. 7.4V in lipo is equivalent to 9.6V nimh ... 15C will not tear your gearbox and gears apart like an 11.1V , that's why most often, 7.4V 15C batteries priced like an 11.1V 20-35C battery.

No.

Read the update I JUST added.

The reason 15C packs are priced like 25C packs is because it is labeled as "airsoft".

I can get a 1300mAh 15C pack for under 15$ from hobby websites.

http://www.hobbycity.com/hobbycity/s...Ah_3S1P_15-25C

It will never be more gentle on your internals. In fact, it will most likely damage the motor and accelerate the wear on it.

Because if the motor is allowed to accelerate very fast, the current drops almost instantly. Where when you limit it, the motor will drain the maximum for a longer period of time before dropping.

Think about it this way. 35A for 1 second then 5A for 9 seconds or 15A for 5 seconds then 5A for 5 seconds.

If you want to save your internals, do a regular maintenance and adjust your mechbox properly. If you are still unsafe, reduce the voltage of your battery.

I have run STOCK TM guns, except for the bushings and spring at 400 fps on 11.1v LiPo for YEARS. EVERYTHING is holding.
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Old December 22nd, 2009, 07:59   #66
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Hey I have a small question.

I just got my charger yesterday, and I was wondering how I should charge the batteries.

I got this charger:
http://www.airsoftparts.ca/store2/in...oducts_id=1030

it can charge at 0.9A or 1.8A.

the battery I am getting is a G&P(yeah I ordered a tamiya to deans adaptor since the charger dont have deans plugs...), 8.4v 2200mAh. Should I charge it on .9 or on 1.8A? also if I undestand the first post correctly, lets say I charge it at .9A, it should take 2.4 hours to charge right? 2200 / 900 = 2.44...
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Old December 24th, 2009, 18:44   #67
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It depends on your battery.

The safe thing to do is to put it on 0.9A setting.

If your pack is made out of AA cells, stick to lower setting. If it is made of 2/3 A or 4/5A, you can put is at 1.8A.
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Old December 25th, 2009, 09:20   #68
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Can you charge lipos with a universal smart NPT charger like the one sold by airsoftparts.ca? Because according to this you just need an additional balancing chip? What are the cons of not balancing the battery?
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Old December 25th, 2009, 13:10   #69
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No

The charger you linked only works for LiPo use.

If you want to blow your house away, feel free to try it though. My one-time experience sent some metal shrapnel into the wall 74' way.
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Old December 25th, 2009, 15:07   #70
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You do hits some very good points in your original post, Kos. I haven't read through the whole 5 pages but I gotta clarify a few inaccuracies in your original post.


1)

Your definitions of energy and power are misleading/inaccurate. You say current is energy and voltage is power. Most people won't even understand the true definitions anyways.


2)

You also state that motors draw as much current as they need, and this isn't totally accurate. We're dealing with DC motors NOT AC motors. AC motors will draw as much current as they need to reach their rated Speed - this is due to AC power being sinusoidal at 60Hz (50Hz in Europe). DC power is not sinusoidal so if you:

- Use the same type of battery cells
- Increase the voltage (ie. add one more cell)
= You'll push MORE current into the AEG DC motor.

Here's some testing done by an airsoft player proving this:
http://www.airsoftcanada.com/showthread.php?t=57571

Quote:
7.2V 3300MaH Sub C battery - 72 Watts @ approx 10 amps
8.4V 3700MaH Sub C battery - 96 Watts @ approx 11 amps
9.6V 1700MaH RIS battery - 92 Watts @ approx 9.5 amps
Note the 1700mah mini battery can only push 9.5 amps. The previous two Sub C packs have similar discharge characteristics and prove my point.

If you look at the "Motor Current Vs Load" graph here ( http://www.girr.org/girr/tips/tips5/motor_tips.html ), it also shows that as you increase voltage (loaded) you'll push slightly more current.

This leads me to the point you make:



3)

Quote:
The second thing that has to be considered, is that the higher the voltage, the more heat the wires will have to support and the more chance there is that the trigger contacts arc.
Heat is more related to the current flowing through the wires, not the voltage. It's like fluid flowing through a pipe, the faster it flows, the more friction is generated (thus more heat). This is why when you spec wires for a system you generally use the current going through them (http://www.interfacebus.com/Copper_Wire_AWG_SIze.html). Although more voltage does mean more current in our case.

You're bang on with the arcing issue and higher voltages, though.



EDIT: I'm a Mechanical Engineer, BTW, but am trying to unlock the mysteries of electricity for myself. The guys I work with are very helpful.
EDIT2: Forgot a link.
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Old December 25th, 2009, 17:06   #71
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Thank you for your inputs.

I know my explanations are not the most accurate in an engineering point of view, but it I worded them so that the average person understands. I did make a mistake however, as voltage is equivalent to potential, where power is the 2-dimension unit representing the combination of both current and voltage.

In theory, DC motors do not drain an infinite amount of current. Every motor can be tested at a stall load to get the maximum current drained into the system.

That value is way past anything that the components of the AEG can hold for more than a few seconds. For all purposes and intend, the value is too large to be relevent. In this perspective, 100A in a system that can take at most 40A can be considered infinite.

When the system is powered by a battery that can deliver around 40-50A peak, it also can be considered infinite in the way that it is a lot above the physical limit of the components.

You cannot push more current into an electrical system because current is a 1-dimension unit. It represents the instantaneous amount of energy that flows in a device. To be able to "push" more current, it would need to be at least 2-dimensional. Voltage is a speed at which the current is moving, thus it can be "forced" faster into the device. The increase in current is only a side effect of the increase in power with the same load or resistance on the circuit.

Also, when using a microcontrolled MOSFET unit such as the Trigger Master, the signal going to the motor is an AC wave form generated by the PWM signal from the controller amplified by the MOSFET.
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Old December 26th, 2009, 05:06   #72
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**WARNING** this is bound to get nerd'ed up here fast...

Kos, can you reword your above post to reflect which portions of my post you are talking about; I'm having a tough time following you.

Quote:
I did make a mistake however, as voltage is equivalent to potential, where power is the 2-dimension unit representing the combination of both current and voltage.
Potential IS voltage, not equivalent to. The terms are use interchangeably, potential just makes more 'sense'. For instance, you rub your feet on the carpet and now have a static charge (voltage/potential)...you now have the 'potential' to shock the hell out of someone - assuming the other person is at a lower voltage/potential than you are. I haven't a clue what you mean by "2-dimension unit"...are you talking about scalars vs vectors???

Quote:
In theory, DC motors do not drain an infinite amount of current. Every motor can be tested at a stall load to get the maximum current drained into the system.

That value is way past anything that the components of the AEG can hold for more than a few seconds. For all purposes and intend, the value is too large to be relevent. In this perspective, 100A in a system that can take at most 40A can be considered infinite.

When the system is powered by a battery that can deliver around 40-50A peak, it also can be considered infinite in the way that it is a lot above the physical limit of the components.
Not sure what you were trying to prove/disprove there.


Quote:
You cannot push more current into an electrical system because current is a 1-dimension unit.
Again, you lost me there.

Quote:
It represents the instantaneous amount of energy that flows in a device.
It's the amount of Coulombs that flow (not necessarily in a device) -> Amperes = Coulomb / second. Volt x Coulomb = Joule (joules is the unit of energy)

Quote:
Voltage is a speed at which the current is moving, thus it can be "forced" faster into the device.
Sorry, but that makes no sense. The mechanical analogy to Voltage is Pressure (ie. PSI, kPa), and current is the fluid flow (ie. GPM, CFM). You can think of an electrical system exactly like a hydraulic system. Mathematically you also model both systems the same way (treat pressure like voltage and current like fluid flow).


Quote:
The increase in current is only a side effect of the increase in power with the same load or resistance on the circuit.
Again, not sure what you meant here but I'm going to try and clarify things.

Power = Voltage x Current

Resistance of the system stays the same (ie. you don't upgrade/downgrade your AEG spring or change anything).

Now if you assume your theory that the motor only draws as much current as it NEEDS, then the power consumption would remain constant. So if this were true, if you increased the voltage you would see the current DECREASE to compensate to keep the power constant.

However it's proven that this isn't the case with our DC AEG motors. When you increase the voltage, the current also increases (thus more power). This is why we get higher ROF when we increase the voltage - assuming the cells have similar discharge characteristics.




Think of an AEG electrical system like an air powered angle grinder. If you want the grinder to spin faster, you need more power...so you need more air! To get more air you either:

a) increase the pressure (voltage) or
b) decrease the resistance - like use a larger diameter air hose - so more air (current) can flow.
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Last edited by Flatlander; December 26th, 2009 at 05:25..
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Old December 26th, 2009, 16:43   #73
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is dead, thanks for all the tips and general FAQ knowledge.
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Old December 31st, 2009, 00:51   #74
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I had a 9.6v 1700mah NiCd, which I did not discharge, sitting away for 2-3 years. It seems to charge fine and is usable once again. However I'm not sure how the capacity held up since I don't usually unload ~1700 rounds.
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Old January 23rd, 2010, 14:46   #75
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Typo.

Should read:

http://www.horizonhobby.com

NiCds are ok to be re-used after 2-3 years, but capacity will suffer.

You will probably notice after a few charges that the battery struggle to shoot 500 BBs... that is because it is getting too old.
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