Had a discussion with a fellow worker the other day. Running of a generator, the skil saw was taking some time to wind up. 50′ 12 ga. supply, with a 3 way and another 50′ 12 ga. cord. I don’t think that is an excessive amount of length and ga. to run one tool. We were commenting on the matter, and he said that there is probably 300′ of cord total off of the generator (some cords were vacant, others were plugged in to tools but not on at the moment). His claim was that, regardless of whether a cord is supplying a running tool or not, all those cords add to voltage drop. I think that the electrons will run from positive to ground in the most direct route. Who is correct?
Jon Blakemore
Replies
voltage drop in a cord depends only on the tools actually running at the time and of course the length of the cord
your slow windup time may be dependent on the generator too, it senses a load and powers up, all takes a few seconds
most electric motors take far more current to start up than run even under the heaviest load
used to use a craftsman tablesaw years ago that actually would not start with a 50 foot14ga cable
had to start by spinning the blade by hand!!! my high school shop teacher im sure would cringe and cancel my diploma
I have the same kind of thing with the new 2 HP motor on my Unisaw. If I just hit the switch, it'll blow a 30 amp fuse, so I had to make a starting cam, like on a hand crank car, and I spin it up with a drill motor to keep the surge down.
-- J.S.
Put 60 amp fuses on it. Fuse wont blow anymore.
i hope that was a joke mark!!
Spinning up a unisaw with a drill is a joke too. Yeah I was jokin'.
OK, here's something somewhat unrelated, but a true story that will chill your blood...
I worked in big old-growth Doug Fir sawmills for a number of years. The primary breakdown saws were 10' double cut bandmills where the logs were canted up and the cants were transferred to the next machine...an "edger" which contained a gang of large circular saws which could be independently set to slice up a cant up to 12" X 72" X 40'.
The arbor shaft in such an edger is massive...6" dia. and about 8' long and powered by 200+ HP electric motor. The arbor is supported by and spins in large roller bearings. The bearing on the end away from the motor is mounted on a "door" and the whole assembly is held in place by a large arbor nut (3" left-hand threads).
To change the saws, the arbor nut is usually removed with a big socket on a rattle gun, but when the saw change is complete, the arbor nut is usually tightened manually with the same socket on a 1" Proto breaker bar.
One morning as the mill was starting the first shift, the filers had completed the edger saw change and had tightened the arbor nut....but neglected to remove the socket and breaker bar which were left hanging on the nut.
The sawmill foreman, a rather portly gentleman, was standing on a catwalk directly above the edger. When the operator started the machine, the breaker bar flew off at high velocity and buried itself clear past the length of the handle knurling (about 6") in the foremans chest. He went down like he'd been pole-axed, suffered a collapsed lung and several broken ribs.
Happily, he survived, but that breaker bar with the bloody handle hung in the office untill the mill closed a few years later. (It was stated by his doctor that a man with less "padding" would not have survived).
Edited 3/6/2003 12:55:16 AM ET by Notchman
OUCH!View ImageGo Jayhawks
That is exactly how a T-handled wrench flies off the chuck in a metal lathe. The first rule of safety is that it is never left in the chuck, ever.
Well, they don't make 60 amp screw-in fuses.... ;-)
The reason for this is that until I get cut over to the new service, all I have is a 30 amp two wire 120 volt services that's been there since 1926. This drill motor and cam as pony motor arrangement is amusing, but for now it gets the job done.
-- J.S.
John
They make 60amp pennies <G>.
John, is that the motor discussed on a previous thread, the 120/240 model you have to run on 120V?
Yes, it's the same one.
-- J.S.
Only possible "in-code" help I could suggest for your 2 HP motor would be to put a 3-5 uF AC run type capacitor across the line just after the switch, that will help reduce the current during startup (and running) by improving the PF - if you want to try it and have a box to put it in (about 4" by 6" by 4"), e-mail me and I'll mail you a couple of surplus ones to try.
Of course, a penny under the fuse is still unacceptable <G>.
Thanks. I finally found my box of big capacitors over the weekend, the smallest I have is 10 uF, would that be OK to try? BTW, my BIL is also looking for a 25 uF 370 VAC, which I also don't have in the junk box. Any leads on a good source?
Thanks again --
-- J.S.
Sorry, but your coworker is wrong and steve is right. You've got about 200' of 12 gauge wire (100' out and 100' back), which, at about 2 ohms per 1000', is .4 ohms, plus the resistance of the connections and the saw cord. With a 40A startup wallop (just a WAG at locked-rotor), that's about a 13% voltage drop. And that's assuming the generator can even provide that much current for startup, and that it's running full-speed already. Sounds perfectly reasonable that the saw starts a bit slower than with a shorter cord of the same gauge.
Be seeing you...
40a startup? I thought the max Amperage figures on tools were for startup and high load times. Could a skil saw really draw 40a?
Jon Blakemore
The nameplate amps is at full rated load. The locked rotor (startup) current draw is considerably more, maybe 40 amps or more. Thats why there is a blink on your lights when you click the trigger on the saw.
They don't nameplate locked-rotor codes on universal motors, so 40A was an educated WAG, but I just happen to have certification data sitting on my desk for a 3 hp Marathon (induction) motor, like what's used on a Unisaw. While the full-load current (FLA) is 12.4A at 230V, the locked-rotor current is 93.1A. It would be 186.2A (!) if it could be configured to run on 115V, while only drawing 24.8A under rated load. Induction motors have a kVA code, or locked-rotor code, on the nameplate, which is used to calculate the approximate inrush current.
Code Letter
kVA/hp Range
A
0.00 - 3.14
B
3.15 - 3.54
C
3.55 - 3.99
D
4.00 - 4.49
E
4.50 - 4.99
F
5.00 - 5.59
G
5.60 - 6.29
H
6.30 - 7.09
J
7.10 - 7.99
K
8.00 - 8.99
L
9.00 - 9.99
M
10.00 - 11.19
N
11.20 - 12.49
O
12.50 - 13.99
R
14.00 - 15.99
For single-phase motors, multiply the code letter value x hp x 1000, and divide by nameplate voltage (generally 115V or 230V). So a 1.5 hp motor with "B" in the Code spot on the nameplate would draw 3.15 X 1.5 x 1000 / 115V = 41A minimum on startup, and up to 46A at the high end.
I seem to recall that Delta's contractor saw with the dual-hp motor had much higher code letter(s); two code letters because it's really a 2 hp, nameplated at 1.5 hp for 115V service, but the locked-rotor current works out to be twice as high at the lower voltage (they need the higher code letter at 115V to adjust for the lower nameplate hp, since it's really a 2 hp at either voltage).Be seeing you...