295LikesTamiya mini cooper
03-06-2011 | 07:41 AM
Tech Initiate
Joined: Feb 2011
Posts: 32
Joules First Law
Heat (Q) = Current squared (I^2) x Resistance (R) x Time (T)
"Where Q is the heat generated by a constant current I flowing through a conductor of electrical resistance R, for a time t." Wikipedia
It's really that simple. Short circuit a battery with different gauge wires, the higher resistance (thinner) wires will heat up quicker. Why? Because the higher level of resistance requires the charges to travel faster through the wire, generating more heat.
If there was not a perceptible change in temp - because that's entirely plausible - it's not explained by your logic either. The explanation is that in these cases, the additional resistance has made no real-world difference to the overall efficiency of the circuit so there's no perceptible change in heat output.
When we talk about heat, we keep current constant. Always.
Myth Busted!
03-06-2011 | 07:43 AM
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Joined: Feb 2011
Posts: 32
03-06-2011 | 08:17 AM
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Joined: Feb 2010
Posts: 54
From: UK
03-06-2011 | 08:28 AM
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Joined: Oct 2006
Posts: 2,367
How many amps do you think a silver can pulls in a mini, and for how long. By far the most power is used from a standing stop on a high grip surface. If you are already moving the amperage is much less, and on a lower grip surface, the tires just spin. I have used 16 ga novak or tq wire in mini's and they don't heat up or cause much resistance (resistance is heat, hence a resistor.) If we were talking about low wind brushless motors on a touring car, 13 or 14ga wire is better suited, but in a mini it really doesn't make a difference.
As for connectors, just use a deans plug and be done with it. Less failure, more secure, and last much longer.
Let's get back to mini's.
p.s.: if v=ra and v is the voltage drop, if we keep v the same and increase r, then a is decreased, in other words, it takes less current over a higher ohm resistor to drop the same number of volts.
The table below is the resistance per foot of typical guage wire: (in r/c we rarely use more than 6")
12 0.001588
13 0.002003
14 0.002525
15 0.00318
16 0.00402
17 0.00506
18 0.00639
19 0.00805
20 0.01015
The voltage drop of 12 guage wire over 6" and conducting 20 amps is: .01588v and 16 guage is: .0402v. Now you might think this is significant (it isn't), but in a mini, a lot of extra power is simply transferred into wheel spin. We also assumed 20 amps continuous, which would mean a runtime of 12 minutes for a 4000mah battery. Since mini's can easily run 20 or 30 minutes with a 4000mah pack, the average draw is about 10 amps, which means a resistance of .02v in a 7.4v battery. Do you really think that makes a difference?
As for connectors, just use a deans plug and be done with it. Less failure, more secure, and last much longer.
Let's get back to mini's.
p.s.: if v=ra and v is the voltage drop, if we keep v the same and increase r, then a is decreased, in other words, it takes less current over a higher ohm resistor to drop the same number of volts.
The table below is the resistance per foot of typical guage wire: (in r/c we rarely use more than 6")
12 0.001588
13 0.002003
14 0.002525
15 0.00318
16 0.00402
17 0.00506
18 0.00639
19 0.00805
20 0.01015
The voltage drop of 12 guage wire over 6" and conducting 20 amps is: .01588v and 16 guage is: .0402v. Now you might think this is significant (it isn't), but in a mini, a lot of extra power is simply transferred into wheel spin. We also assumed 20 amps continuous, which would mean a runtime of 12 minutes for a 4000mah battery. Since mini's can easily run 20 or 30 minutes with a 4000mah pack, the average draw is about 10 amps, which means a resistance of .02v in a 7.4v battery. Do you really think that makes a difference?
I'd thought for a long time that the wires being used to wire a Mini were unnecessarily large. I just did it cause everyone else did it and that was the conventional wisdom. I will bet that almost all of you did the same thing. I checked most of the Minis at the track yesterday and all of them were using "big" wire.
To the fella who may have been referring to me as the common denominator in a lot of these "discussions", I'd like to say Thank You. That was a nice compliment although you probably didn't intend it to be. I'll admit that I like "stirrin the pot a little" just to see what's in there. Sometimes there's some really good "stuff" in there and sometimes not. You never know until you "stir the pot" a little. Whenever a discussion with varying points of view occurs, the possibility learning something new also exists.
03-06-2011 | 09:28 AM
If it's in a typical electrical circuit of course it will heat more if you increase the resistance sufficiently and the current draw remains constant (which is how heat is calculated). It has to:
Joules First Law
Heat (Q) = Current squared (I^2) x Resistance (R) x Time (T)
"Where Q is the heat generated by a constant current I flowing through a conductor of electrical resistance R, for a time t." Wikipedia
It's really that simple. Short circuit a battery with different gauge wires, the higher resistance (thinner) wires will heat up quicker. Why? Because the higher level of resistance requires the charges to travel faster through the wire, generating more heat.
If there was not a perceptible change in temp - because that's entirely plausible - it's not explained by your logic either. The explanation is that in these cases, the additional resistance has made no real-world difference to the overall efficiency of the circuit so there's no perceptible change in heat output.
When we talk about heat, we keep current constant. Always.
Myth Busted!
Joules First Law
Heat (Q) = Current squared (I^2) x Resistance (R) x Time (T)
"Where Q is the heat generated by a constant current I flowing through a conductor of electrical resistance R, for a time t." Wikipedia
It's really that simple. Short circuit a battery with different gauge wires, the higher resistance (thinner) wires will heat up quicker. Why? Because the higher level of resistance requires the charges to travel faster through the wire, generating more heat.
If there was not a perceptible change in temp - because that's entirely plausible - it's not explained by your logic either. The explanation is that in these cases, the additional resistance has made no real-world difference to the overall efficiency of the circuit so there's no perceptible change in heat output.
When we talk about heat, we keep current constant. Always.
Myth Busted!
Bulb's fourth law : Keep science out of mini racing.
03-06-2011 | 11:06 AM
Giant wire, like most of the bling is not needed on a racing mini. My M-03 is mostly black now, where a few years ago it was a sea of blinding blue. There's 16ga now where once there was 12ga.
The bonus is that it's a lot easier to work with. I'm club racing, not aiming for a trophy after all.
The bonus is that it's a lot easier to work with. I'm club racing, not aiming for a trophy after all.
03-06-2011 | 12:27 PM
I shall send you a p.m , since you weren't the person I was referring to.
Discussion is good, banging on and on the way two people have, is verging on trolling.
Discussion is good, banging on and on the way two people have, is verging on trolling.
Thanks, Bill. Exactly the answer to the question I posed a few pages back with the info that I wanted to know. Glad that this subject can be brought to close.
I'd thought for a long time that the wires being used to wire a Mini were unnecessarily large. I just did it cause everyone else did it and that was the conventional wisdom. I will bet that almost all of you did the same thing. I checked most of the Minis at the track yesterday and all of them were using "big" wire.
To the fella who may have been referring to me as the common denominator in a lot of these "discussions", I'd like to say Thank You. That was a nice compliment although you probably didn't intend it to be. I'll admit that I like "stirrin the pot a little" just to see what's in there. Sometimes there's some really good "stuff" in there and sometimes not. You never know until you "stir the pot" a little. Whenever a discussion with varying points of view occurs, the possibility learning something new also exists.
I'd thought for a long time that the wires being used to wire a Mini were unnecessarily large. I just did it cause everyone else did it and that was the conventional wisdom. I will bet that almost all of you did the same thing. I checked most of the Minis at the track yesterday and all of them were using "big" wire.
To the fella who may have been referring to me as the common denominator in a lot of these "discussions", I'd like to say Thank You. That was a nice compliment although you probably didn't intend it to be. I'll admit that I like "stirrin the pot a little" just to see what's in there. Sometimes there's some really good "stuff" in there and sometimes not. You never know until you "stir the pot" a little. Whenever a discussion with varying points of view occurs, the possibility learning something new also exists.
03-06-2011 | 02:52 PM
Tech Initiate
Joined: Feb 2011
Posts: 32
I shall pick up a pack of those along with some 5mm adapters.
03-06-2011 | 03:40 PM
Tech Elite
Joined: Feb 2009
Posts: 2,972
From: All over the place
Q=Uexp2*t/R
which is where indeed you can replace Uexp2/R=Iexp2*R to give the equation you wrote above, but this is what is misleading you (and others) into reverse engineering the equation and deciding that heat depends directly proportional to resistance, when in fact is inversely proportional.
That implies assuming that you can choose your current which you can not, as I said before. All you can do is choose your voltage and your load (resistance). Current is a derived measure from these two.
When we talk about R/C all you have is a battery with a fixed voltage so we keep voltage constant. (not sure I understand why you postulated keeping the current constant anyway as even that doesn't help your case at all).
03-06-2011 | 03:54 PM
Tech Initiate
Joined: Feb 2011
Posts: 32
Yep, that's exactly where you're wrong.
Q=Uexp2*t/R
which is where indeed you can replace Uexp2/R=Iexp2*R to give the equation you wrote above, but this is what is misleading you (and others) into reverse engineering the equation and deciding that heat depends directly proportional to resistance, when in fact is inversely proportional.
That implies assuming that you can choose your current which you can not, as I said before. All you can do is choose your voltage and your load (resistance). Current is a derived measure from these two.
When we talk about R/C all you have is a battery with a fixed voltage so we keep voltage constant. (not sure I understand why you postulated keeping the current constant anyway as even that doesn't help your case at all).
Q=Uexp2*t/R
which is where indeed you can replace Uexp2/R=Iexp2*R to give the equation you wrote above, but this is what is misleading you (and others) into reverse engineering the equation and deciding that heat depends directly proportional to resistance, when in fact is inversely proportional.
That implies assuming that you can choose your current which you can not, as I said before. All you can do is choose your voltage and your load (resistance). Current is a derived measure from these two.
When we talk about R/C all you have is a battery with a fixed voltage so we keep voltage constant. (not sure I understand why you postulated keeping the current constant anyway as even that doesn't help your case at all).
If you've got any relevant sources feel free to pm me but I'm not taking the bait (any further) in public.
03-06-2011 | 04:07 PM
Tech Elite
Joined: Feb 2009
Posts: 2,972
From: All over the place
Relevant sources for what? It's all high school physics. You can find any of this anywhere you care to look.
Besides, you don't need any sources, you need to understand math. It's your own equation which I rewrote, using I=U/R, Ohm's law, you know?
And before launching slander, be sure you know what you're talking about. I didn't call you anything or accused you of anything. If you can keep it civil I see no reason to continue elsewhere.
Besides, you don't need any sources, you need to understand math. It's your own equation which I rewrote, using I=U/R, Ohm's law, you know?
And before launching slander, be sure you know what you're talking about. I didn't call you anything or accused you of anything. If you can keep it civil I see no reason to continue elsewhere.
