You are for sure more capable of judging this, i just look simplified at the sled as a pendulum that with 0 friction (unknown variable) swinging to far if speed is increased. My limited view is that acceleration and deceleration planing might take care of this. I don’t know.
You have a good point there, and as i said before, i value your opinion. Thanks for taking the time to answer me.
20RPM for the motors sounds about right from my recollection. Keep in mind though, that the actual maximum linear movement over the work surface is not uniform. I believe I recall writing a mechanism into the firmware that looks at the RPM of each motor for each position over a line and sets the maximum feed rate for that line at a value that does not exceed the motors RPM. I think it has been a long time.
Sorry for not mentioning you as one of the people that can help on this.
Kind regards, Gero
Anyone on this post remembering where it is in the Firmware 1.26? Have trouble finding it to do the RPM test at F1000. Thanks.
sysSettings.maxFeed = 700; // int maxFeed;
calibration movements move at 0.9*maxFeed
default movement speed is maxFeed/2
sys.feedrate = constrain(sys.feedrate, 1, sysSettings.maxFeed); //constrain the maximum feedrate, 35ipm = 900 mmpm
but maxFeed is set at 700, so 900 mmpm is only a comment. we are constrained at 700 as far as I can tell for classic Maslow from 1.28 firmware I’m running. I did not change this value.
Ok, just one more question before i go to bed, what encoder is on this motor?
probablay the same one used now 7ppr.
My math might be wrong, but with the current 20 rpm motor and a sprocket with 10 teeth and #25 chain (6.35 mm/link), the maximum chain speed (regardless of what direction the sled is moving) is 20 rev/min * 63.5 mm/rev = 1270 mm/min. If we limit feedrate at 700 mm/min, then our motors never exceed 11 RPM. The feedrate was limited so the sled could move across the entire surface without excessive error (i.e., sled not keeping up alarm). The worst case scenario is movement around the top middle because the motors are pulling against each other and I think that’s where we overloaded the original motor controllers. Not sure if anyone really has tried to push it with the TLE5206 boards. I think the stock motors can actually handle 24V (I think @bar got that info from the manufacturer at one point)… double the voltage and you can half the current needed.
Also remember that the sled isn’t captured and if you try to push through the wood too fast it might get unstable.
The current is directly proportional to the voltage and inversely proportional to the resistance. Any alteration in the voltage will result in the same alteration of the current . So doubling the voltage will cause the current to be doubled.
assuming steady state and knowing ohm’s law (V = IR). If R is constant and V doubles, then I has to decrease.
(easbay is correct for the up above. The motor does not act as a true resistor, so it does not follow. I just walked right into that one.)
This is a power issue. P=IV. For some power, if voltage doubles, current will have to drop in half for the same power output or the same speed or the same torque in the case of the motor.
I = V/R
If V increases and R is constant then I is higher.
You are increasing the voltage without changing anything else, why “current will have to drop”??, that’s not how it works.
Unfortunately, since R does not change, a higher V will result in higher P (V*I) hence doubling the V you are also increasing P and I
I doubles with double of voltage
If P is constant and V goes up, then I will have to drop.
P = IV
200 W = 100 V * 2 A
200 W = 200 V * 1 A
this is why 220 wires can be smaller for the same power motor. This is why when computer chips go from 5V to 3.3V volts they use more current.
You are correct though about ohms law and I was incorrect that if resistance is constant and V goes up, then I will go up for the same R.
A motor rated at up to 24V, so are we getting that RPM? Thinking about the g-code to test, but neither a straight down nor a horizontal cut. Have to dig back in the “pull tight” to get the max for this test I guess.
this old posts said motors/gearbox go faster and are warmer run at 24v for testing only, the first post above that says 41rpm at 24v, so no we are not getting that rpm .
You are taking R out of the equation.
12v * 0.2A = 2.4W
12V / 0.2A = 60R
now increasing V without changing R
24V / 60R = 0.4A (note double the I at the same R)
24v * 0.4A = 9.6W
I agree to disagree with this statement. Our motors have been able to break a lot of stuff and even where capable to make my bone marrow vibrate during debugging. 2 of them will take care of heavy sleds at ease.
Lol… didn’t mean to start a debate. I was trying to say that with a higher voltage, only half the current is needed for the same feedrate. Therefore, the motor controller has more headroom and maybe the feedrate could be increased.