On our bot, we have the +12 volt wire going from the battery, through the reset-able breaker, to the distribution rail. From there, it goes in one side of a maxi block. A new wire comes out the other side of the maxi block and goes to the small black fuse panel terminal, then a third wires comes off the small fuse panel terminal and goes to our last maxi block.
Are we unknowingly introducing a high amount of transmission resistance by having the multiple connections in our wire run, compared to running individual wires from locations on the distribution block and going directly to each component?
Squirrel is correct, you need to run the red wire from the distribution block to Each Maxi and fuse panel. Each Maxi and fuse panel should have a red wire running back to the distribution block. They can’t go from maxi to the fuse panel to another maxi.
Chief Delphi is so great… in the time it took me to look up the rule (and talk to a few students and such), two other people posted the same point.
In any case, R50 states (in part):
All circuit breaker distribution panels must be connected directly to the power
distribution block. No intermediate connections are permitted.
As to the engineering question at the root of your question, really what we would need to know is how much current was being drawn at each point. If all your drive motors were on the first maxi-block and everything else was low current, then the amount of power lost would be minimal and worrying about the “resistance is futile”. But it really is against the rules, so this was mostly a cheap excuse to make a borg quote.
Jason
Tom,
There is a reason behind the First Electrical Wiring Diagram and it will be obvious if you think about it. Each panel that is in a daisy chain adds to the current that flows through the next in line. In thinking about Ohm’s Law, this increase in current causes an increasing and significant increase in the voltage drop of the wire between the panels.
I use a term, the Wire Foot, when describing these losses. A wire foot is the resistance of a #10 wire (.001 ohm/foot) and the current passing through a drive motor (~100 amps under load). Under these conditions, every foot of wire will drop 0.1 volts. Some of the other losses become significant as you start to think about real world robot loads. A four motor drive in stall can easily draw over 400 amps from the battery. When you compute the losses with the battery internal resistance you start to draw down the available buss voltage for all robot systems. This is a critical issue for the RC which drops out at less than 8 volts. So you can see that every inch of wire between the RC and the battery that has significant current flowing through it also reduces the voltage to the RC.
Battery Internal Resistance=11WF
#6=0.5WF
#12=~2WF
Victor=6-8WF
Poor screw terminal=1-3WF
Sum total of breakers=1-3WF
Bad Crimp=1-3WF
In your original wiring, the losses at a four motor stall could be…
11WF(battery)+2WF(4’ of #6)+6WF(total of all losses from connections)=19WF x 4 (400 amps) x 0.1volt/WF = 7.6 volts or only about 5 volts available for the RC with the battery fully charged. Since the RC cuts off at 8 volts this condition would only exist during the time the RC was actually in control.
However, if wired as in the First diagram the losses become significantly less since the RC does not derive power after the maxi block, the losses in that wiring do not affect the RC power. Your only losses come from the battery cable and the internal battery resistance that still share the motor current. Even the losses due to connections are reduced allowing the RC to remain near or above the critical cutout voltage.