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Old 11-07-2007, 01:02 PM   #16
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LOL...then make sure you buy another super-trick option part to offset the weight.
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Old 11-07-2007, 01:06 PM   #17
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LOL

I have my limits. 12GA is big enough, you won't find any auto starter or battery cables in my RC car.
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Old 11-07-2007, 02:15 PM   #18
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Originally Posted by CarKing View Post
Wow this turned out to be more of a debate than I thought it would. What do you guys use personally? Does anyone have a GTX that they use the 12ga wire in that they could give me some tips on when I rewire it? Those hole on the esc are a pain to get the wires in once they have had any solder in them at all!!!
What I do is ensure the stripped wire end is tinned with solder and the holes on the ESC are lightly covered completely with solder. I then turn the GTX upside down and secure it to the edge of a table with servo tape (I put it on the heatsink). then hold on to each wire with some needle-nosed pliers and heat the top of the hole on the ESC while pushing the wire through the hole with the pliers. I find this works great as it gets the wire inserted right up to the heatshrink to eliminate the risk of shorting out your speedo. I always run an xacto knife between the holes to ensure no solder is touching two holes. Hope this helps.
-Kane
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Old 11-07-2007, 02:20 PM   #19
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I use a tool (that came with a Weller soldering gun kit) to open the holes. It is a tapered piece of aluminum, like a really big toothpick. I heat the solder and put this in the hole. It keeps the solder clear of the hole and you don't need to worry about spatter. The solder doesn't stick to it since it is aluminum

Another trick if the holes arean't quite big enough for 12Ga wire: I tin the end of the wire going into the ESC board holes and cut it to the proper length after it is tinned. Then I kind of grind it down (uniformly) by rolling the bare/tinned part against a cutoff wheel in a Dremel until it is small enough to fit. Yes, you are making the wire smaller for a length of about 1/32 of an inch, but the rest is still 12Ga.
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Old 11-07-2007, 02:46 PM   #20
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That is patently untrue. I have 14 guage wire connected to a Competition Electronics single cell holder I use for rematching cells. I discharge the cells as 35 amps for up to 400+ seconds. The wire gets warm, but nowhere near hot enough to melt the solder connections.

I am prety sure from my schooling, (its been a long time )

but the same amps at more voltage disipates more watts

using ohms law for Power (watts)

P = V x I (P for Power (watts)) V for Voltage , I for Amps.

so posibly pulling say 35amps at 7.2volt will get hotter than at 1.2volts (1 cell)

try the formula out using a different volatge but keep the other vairables the same.


example 1- 7.2volts 20amps = 144 watts

example 2- 1.2volts 35amps = 42 watts


(not that you were not saying that it wont, but just some food for thought)


I also did track testing times, comparing 12awg to 14awg with 27turn and did not notice a diff at all.

Cheers.....
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Old 11-08-2007, 12:31 AM   #21
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Nope, your formula is correct, but you are thinking wrong.

The voltage doesn't matter at the motor or at source (the batt), it's the voltage drop you loose on the wire which makes the calculation !

If you have for example 100 volts, with a cable with .1 ohms and a current of 10A attached to an motor. The wire will U (volt) = I (current) x R (resistance) get 10Ax.1=1volt ! The motor will get 99 volts.
The power in the cable will then be 1voltx1A=1Watts

Now the same cable 0.1 ohms, same 10A current, but with a 5volt Batt.
The wire will U (volt) = I (current) x R (resistance) get 10Ax.1=1volt ! The motor will get 4.0 volts.
The power in the cable will be the same 1voltx1A=1Watts

12awg or 14awg have much lower resistance, this is just an calculation example. But you can see easily that lower resistance has a big gain with low voltages and high currents !
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Old 11-08-2007, 12:51 AM   #22
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yeah cool, my post was mainly only in regards to how hot a 14awg wire would get.

at X amount of amps, with different voltages

assuming 1.2volts at 20amps / compared to 7.2volts at 20amps

would the wire be the same temp?

i think it would be hotter at the higher voltage as there would be more voltage drop at the higher voltage.

I assume we are both correct, I was just talking off topic a bit.

(when i do a discharge with my ICE of 10amps at 1 cell the ICE gets warm,

when i do a discharge with my ICE of 10amps at 6 cell it gets way hotter)
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Old 11-08-2007, 02:19 AM   #23
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Huh?

You guys all seem to be hitting the same pipe...LOL

You can't figure out what the wire will drop without figuring what the load (the motor) drops. If you guys are going to go looking for fairies, you need to locate the elephant first.

If you have a 1ohm load and .0015 ohm/foot wire (nominal resistance for solid 12g wire), versus the same load with .0025 ohm/foot wire (14g solid wire), you have to figure out the total load (motor resistance + additional drops) to figure the current...THEN you can get to the amount of voltage you lose in the wire.

Using the nominal figures above and a 1ohm motor, I get a .0063A difference for one foot of solid wire between 14 and 16 @ 7V. High-grade stranded is MUCH much much better.

For kicks, with 70V, it's .063A or ten times the difference for ten times the voltage. Same resistance with less voltage = current drops by less.

Honestly...if you can "feel" the difference in wire gauge...why can't you "feel" the difference in wire weight as well? The additional weight would most assuredly cancel out any additional efficiency in a high-resistance (high-wind) application.


Quote:
Originally Posted by snoebel View Post
Nope, your formula is correct, but you are thinking wrong.

The voltage doesn't matter at the motor or at source (the batt), it's the voltage drop you loose on the wire which makes the calculation !

If you have for example 100 volts, with a cable with .1 ohms and a current of 10A attached to an motor. The wire will U (volt) = I (current) x R (resistance) get 10Ax.1=1volt ! The motor will get 99 volts.
The power in the cable will then be 1voltx1A=1Watts

Now the same cable 0.1 ohms, same 10A current, but with a 5volt Batt.
The wire will U (volt) = I (current) x R (resistance) get 10Ax.1=1volt ! The motor will get 4.0 volts.
The power in the cable will be the same 1voltx1A=1Watts

12awg or 14awg have much lower resistance, this is just an calculation example. But you can see easily that lower resistance has a big gain with low voltages and high currents !
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Old 11-08-2007, 05:26 AM   #24
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I have always used 14ga wire.
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Old 11-09-2007, 09:46 AM   #25
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Quote:
Originally Posted by snoebel View Post
Nope, your formula is correct, but you are thinking wrong.

The voltage doesn't matter at the motor or at source (the batt), it's the voltage drop you loose on the wire which makes the calculation !

If you have for example 100 volts, with a cable with .1 ohms and a current of 10A attached to an motor. The wire will U (volt) = I (current) x R (resistance) get 10Ax.1=1volt ! The motor will get 99 volts.
The power in the cable will then be 1voltx1A=1Watts

Now the same cable 0.1 ohms, same 10A current, but with a 5volt Batt.
The wire will U (volt) = I (current) x R (resistance) get 10Ax.1=1volt ! The motor will get 4.0 volts.
The power in the cable will be the same 1voltx1A=1Watts

12awg or 14awg have much lower resistance, this is just an calculation example. But you can see easily that lower resistance has a big gain with low voltages and high currents !

Sorry I have never seen .1 ohm wire except in a hot wire cutter.
Go Back to School and get a degree in EE before posting .

My $.02



Quote:
Originally Posted by Turbo Joe View Post
Huh?

You guys all seem to be hitting the same pipe...LOL

You can't figure out what the wire will drop without figuring what the load (the motor) drops. If you guys are going to go looking for fairies, you need to locate the elephant first.

If you have a 1ohm load and .0015 ohm/foot wire (nominal resistance for solid 12g wire), versus the same load with .0025 ohm/foot wire (14g solid wire), you have to figure out the total load (motor resistance + additional drops) to figure the current...THEN you can get to the amount of voltage you lose in the wire.

Using the nominal figures above and a 1ohm motor, I get a .0063A difference for one foot of solid wire between 14 and 16 @ 7V. High-grade stranded is MUCH much much better.

For kicks, with 70V, it's .063A or ten times the difference for ten times the voltage. Same resistance with less voltage = current drops by less.

Honestly...if you can "feel" the difference in wire gauge...why can't you "feel" the difference in wire weight as well? The additional weight would most assuredly cancel out any additional efficiency in a high-resistance (high-wind) application.

I use 14 and 16 gauge in stock and 19T because it is lighter and you will never feel the difference in the power loss from the wire at 3 to 6".

Food for thought the wire size inside the speedo is MUCH SMALLER and the wire inside the FETS is almoset the size of a hair but only .1" long
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Old 11-09-2007, 12:34 PM   #26
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I know i'm gonna get flamed here, but with the low IR of today's batt and ESC, won't you try to get rid of the bottle neck from the wires? If you had a choice, would you mount the ESC close or as far from the motor as possible?

With the M03, i see ppl cutting away into the center of the chassis to mount the esc as closest to the motor as possible(14ga). And they win. I think, especially in stock racing, everybit counts.
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Old 11-09-2007, 08:09 PM   #27
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Originally Posted by redbones View Post
I know i'm gonna get flamed here, but with the low IR of today's batt and ESC, won't you try to get rid of the bottle neck from the wires? If you had a choice, would you mount the ESC close or as far from the motor as possible?

With the M03, i see ppl cutting away into the center of the chassis to mount the esc as closest to the motor as possible(14ga). And they win. I think, especially in stock racing, everybit counts.
That's because the only other place to mount it is on top. That and trying to keep the wires short for weight more than resistance.

You don't get rid of any bottleneck. The biggest (by far) "bottleneck" is the motor. The wires have nearly insignificant resistance and changing to twice as heavy a wire to save (a nearly immeasurable portion of) the wire resistance is silly for stock. I think the term that applies is "pissing into the ocean".
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Old 11-10-2007, 03:25 AM   #28
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So the bottom line is,

IT DOESN'T MATTER.

Just get rid of those Tamiya connectors(3mΩ~35mΩ) that the speedo comes with, and you are set. Worry more about heat management. That has a bigger factor to power output and overall resistance.
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