nauticom

I have a 1/6 scale airplane I am finishing up and am curious about using some sense of math or logic in calculating battery load for the following configuration:
Seven servo's(7) total: throttle, elev, rudder, 2 ail, 2 flap
Servo's are JR 531-these are analog
Currently I have a 1100 mahr nicad.
Any help in pointing me to the right direction would be helpful. I don't want to leave anything to chance.
Thanks,
Bob Masterson

IFLY3W

My Feedback: (5)

An 1100 NiCad will get you by..... But not for long....... 7 Analog servos, I would use either NiCad 2400 or a NiMH 2700. EIther 4.8V or 6V is fine. With 6V, all you'll be doing is increasing servo torque and speed. Either choice is a good one. Or if you want a little battery redundancy, you could go with a pair of 1650's with 2 switches, or 1 battery and 2 switches.... Options are endless there... But battery choices are of what I would choose!!

Good Luck!!

Chris~

mglavin

My Feedback: (31)

You'll likely consume approximately 15mA per minute of flight time.

60% usable capacity of your flight pack calc's to 1100/60%=660ma

660mA/15mA=44, or 44 minutes of estimated flight time.

Keep in mind there is NO substitute for a thorough pre-flite which would include a loaded voltage test with an expanded scale voltmeter. These meters are readily available and place a known load on the battery and display the voltage under same. Specifically your placing a 500-1500mA load on the flight pack and viewing the voltage under load. This is a reasonable method but not the best method to derive the condition and or state of the battery pack.

Typical NO FLY voltages are: 4-cell=4.4V, 5-cell 5.5V.

JPMacG

My Feedback: (2)

I am in a similar situation. I am not so concerned with choosing a battery with enough mAh - I understand all about that. What has me a little confused is the following;

Should I be worrying about the size of the batteries with regard to the internal resisitance of the cells? For example a pack made of cells with low internal resistances would be able to supply higher peak current than a pack made of cells with higher resistance.

I understand that physically larger NiCds have lower internal resistance, for example an AA pack has lower resistance than a AAA pack. Also I understand that lower capacity packs have lower internal resistance than higher capacity packs of the same size. That is, for example, a 700 mAh AA pack would provide more peak current than a 1100 mAh AA pack (but of course would not fly as long).

Should I think of using C cell packs for my 6 and 7 cell airplanes? Do you guys ever worry about this?

PilotLight

My Feedback: (1)

How did you compute 15Ma per minute? What about 6 servos?

mglavin

My Feedback: (31)

Every specific cell and technology carries it's own IR factor, NiCad offers the least, NiMH and then Lithium is the highest to date.

For our flight pack purposes the peak demands on sport aircraft are within the limits of AA packs IMO (600mAh or larger). Capacity is a factor with this in mind. It all boils down to the needs of the specific model. That said we generally associate 'C' or capacity with the ability of the cell in hand to discharge at a specific 'C' value. For example Lithiums range anywhere from C2-C20 at this time (capacity in mA multiplied by the stated 'C' value). A 2200mAh Lithium-Ion cell is therefore able to discharge at sustained voltage levels of 4.4A ) 2200/1000=2.2, 2.2 x C2 = 4.4A. Same holds true for all cell technology, but you need to know the 'C' value or factor it based on the cells IR or rely on the battery manufacturer/packager to provide an amperage rating for the specific cells in play.

mglavin

My Feedback: (31)

15mA was derived from flight test evaluations. To be honest the 15mA is likely a little high for the seven servo load mentioned and analog verses digital servos skew the comparison somewhat (digitals consume more power). Its pretty straight forward to accomplish a baseline for your model.

You'll need a charger/discharger device that displays the mA discharged.

1) Evaluate your battery by cycling at least three times through the charger/discharger. This will provide a baseline for the batteries capacity. Lets say an 1100mAh battery delivers rated capacity.

2) Fly the model for a fixed amount of time in a manner consistent with your normal routine. Maybe 30 minutes.

3) Discharge the battery thereafter to the same cutt-off voltage (.9V per cell is typical for NiCD-NiMH) previously used and do the math... Lets say the battery provided and additional 500mA.

1100-500=600. This tells me we used 600mA. 30 minutes flight time divided by 600mA = 20ma per minute consumption.

The thing is the battery will not provide the required voltage level until depletion, therefore I use 60% of the rated capacity as a guideline, anything past this value usually falls off the chart rapidly and is unable to provide current/voltage under load. I have graphed the discharge of many flight packs and found this number to be a safe value.

1100mAh at 60% = 660mAh thats safely usable.

Red B.

JPMacG wrote: In my experience the internal resistance of modern NiCd and NiMh batteries is not an issue for normal radio applications. The combined resistance of switch harnesses, servo connectors and wires are in my experience much higher than the internal resistance of the battery itself.


mglavin wrote Your calculation is not entirely correct.
IF, the current draw was 15 mA, a 1100 mAh battery with 60% efficency would last for 60%*1100 mAh/15 mA=44 HOURS, not minutes!

On the average the current will be much higher than 15 mA during normal flying conditions. The no-load current of a typical servo is approximately 200 mA and depending on the load it can be much higher.

It is almost impossible to guesstimate the current during flight as too many variables has to be taken into account.
As a rough guide the old standard, 4 servos and a 5-600 mAh battery should point you in the right direction.

If using a 6V system instead of a 4.8V system I would recommend a 50% increase in battery capacity, because the current draw will increase by roughly 50%.

If digital servos are used instead af analog ones I would increase the capacity by 50%

Another factor to take into account is wether dual aileron servos OR/AND elevator servos are used. If OR I would increase the battery capacity by 50%, if AND I would double the battery capacity.

To sum it up: Nauticom, for your seven analog servo's: throttle, elev, rudder, 2 ail, 2 flap I would go ahead and use the 1100 mAh battery presuming that the flaps are used only now and then.

Disclaimer: The reccomendations above works for me. Your mileage may be different.

/Red B.

ini

Let's be more precise, all.

The above consumption should be 15 mAh per 1 minutes flying (instead of 15 mA per minute).
In real life that means current of 900 mA = 0.9 A (15 mA * 60 min = 15 mAh).

I have measured 120 - 160 mAh energy comsumption during 10-15 minutes flying of 2 meter pattern plane with 6 servos (analog). Since I went to digital servos I have used Lithium battery which I monitor using voltage, not mAh:s .

So, I would suggest using ESV, of course, but also a computer controlled field charger. 1100 mAh will be enough for couple of flights. After that charge and check how much was consumed during those flights. Or better yet, dicharge the pack and see how much was still available.

Just my two cents,

ini

mglavin

My Feedback: (31)

The math I provide is correct, 15mA per minute is the average consumption of the model in my example. 660mAh/15mA per minute flight time = 44 minutes... Unless I'm missing something.

Your right here, that's why we cycle and grade the battery in question and subsequently perform a timed flight test evaluation, discharge the battery used in the flight test and garner real world data from the model in question...

This seems a bit extreme and is well outside my findings, have you evaluated a flight system with 4.8 and 6.0V to arrive at this assertion?

This makes sense, but is likely a little overboard IMO

I concur here. I find that an 1100mAh battery will easily provide the required power for a model as depicted in the above examples, I've been doing it for many years now.

mglavin

My Feedback: (31)

So your suggesting 900mA is consumed in a 60 minute time frame. All right then that works out to 15mA x 44 minutes = 660mA.

(15 mA * 60 min = 15 mAh) ??? Did you mean 15mA * 60(h) = 900mAh


Your findings are similar to mine here...

Red B.

mglavin wrote: You are missing something. Battery capacity is current * time, NOT current / time as you suggest.

If the current draw is 15 mA during a 1 minute period (1/60 hour), you will consume 15 mA * (1/60) h = 0.25 mAh.
The battery will last for 660 mAh / 0.25 mAh/min = 2640 minutes = 44 hours.

Its much easier to do the calculations if hours are used as the unit of time:
Again, asuming the current to be 15 mA the battery will last for 660 mAh / 15 mA = 44 h.

If your battery lasts for 44 minutes, than the average current has been 660 mAh / (44/60 h) = 900 mA as INI previously wrote.

In reply to my message:
mglavin wrote:
O.K. I measured my Hitec 5295 MG servos and under no load conditions the current jumps from approx 200 mA to 250 mA when switching from 4.8 V to 6 V operation. That's a 25% increase in current. Because I haven't performed current measurements with a load I choose to err on the safe side.

/Red B.

Geistware

My Feedback: (16)

What I do is turn on my servo exerciser on my 9C and run the pack until dead.
This will let me calculate the average current draw of the system.
From this, I allow for 6 flights or a little over one hour of on time.

mglavin

My Feedback: (31)

Red/INI

The math is making sense NOW, thanks for the schooling.

On the 50% capacity increase thing.

If an amount of energy to perform a known amount of work is a constant. Can we assume the increased speed (time) adsorbs the increased current? Therefore validating the increased capacity/current conundrum. 25% increase in current seems like a hefty price to pay for a slight increase in speed.

Red B.

You bet :-).

It seems that you are right about the speed versus current. I looked up the spec sheet for my HS-5925 servos and it appears that when increasing the voltage from 4.8 to 6 V, the no load current increases from 190 mA to 240 mA (almost the same as my measurement I'm glad to say), i.e. a 26% increase in current and thus also in power. The 60 deg. transit time decrease from 0.10 to 0.08 s, i.e. a 20% reduction.
Doing the same calculations for HS-5475 servos a 12% increase in current (160 -> 180 mA) is accompanied by a 22% decrease in transit time (0.23 -> 0.18 s).
So, it seems that you are right, under no load conditions the increase in current is compensated for by the decrease in transit time, i.e. the same amount of energy (work) is used to move the servo no matter what the voltage. Very interesting. Thank you for pointing this out!
Now, what will happen when the servo is working against some load? I think some measurements are called for.

/Red B.

ini

From theoretical point of view increase of 25% in voltage causes 56% increase in current consumption under load.

Basically, power that a servo or any other device outputs is force times speed. Increased voltage will increase both force and speed, so increase in capacity will be 1.25^2 -> +56% .

In real life most of the time servo is not that much loaded, so idle consumption is more inportant. In practise going from 4.8 to 6 V increases comsumption by 25 %.

Going digital is a more complex thing. If only slight loads are applies there is not necessarily any change in consumption. Heavy loads on servos as in 3D could cause double or quadruple amperage. My own experience with 2m pattern planes is in ball park of 50% more.

ini