Shopsmith Forums

I am rebuilding a 3/4 hp AO Smith from a 1954 Greenie. I need a replacement start up capacitor. Not sure what type to get as the markings on the old one are rubbed off. My multimeter testing indicates it is bad. Any help is appreciated. It needs a capacitor housing and some other parts in addition to new bearings. The motor is one I have never seen either in Jacob Anderson's old site or Shopsmith 505 repair site. Duckman

Suggest you supply a close-up photo of the nameplate, or recite the specs on the nameplate.

My guess is that any capactior from any single-phase 3/4 HP motor would be OK, but let's see if anyone here can provide specifics.

How did you test it? Start capacitors rarely go bad. If the motor does not start it is more likely the start switch. What are the symptoms? The replacement value is not critical. The voltage rating should be 125 V or higher. The capacitance is not critical. 400 µF +-50 should be fine. It is an electrolytic capacitor and needs to be the same form and physical size to fit.

a close up of the tag would help , for the exact match there should be a model number on the tag and you should be able do a search on the web to match the capacitor to the model or find a rating tag that is readable . JMO if it is too bad you should call HQ and they should be able to help .

A local electric motor repair shop can tell you what capacitor you need, and they will also have one on hand to sell you. That would be the easiest way to get it back and running.

Go to the sticky post at the top of the Maintenance and Repair forum, Shopsmith Large Format Drawings, Illustrations and More
In the first post click on the red "My Google Drive".
Click on the folder "Motors" folder
Click on (download) the file "How to Replace AO Smith Motor Parts.pdf"

Here is the calculation for the capacitance required to start the motor from the inrush current state(or, locked rotor state).

C(uf) = (p*n*1000) / (v*v*f)
Where P is the power of the motor in watts, n is the motor efficiency, v is the voltage, and f is the frequency.

Example: C = (746*80*1000) / (220*220*60) 746 watts, 80 percent eff(safe guess), 220VAC, 60Hz

C = 59,680,000 / 2,904,000

C = 20.55 uF. Will be safe with a 25uF cap, usually get a bit over working voltage, like 300.

C(uf) = (P x n x 1000) / (V² x F)

WTH did that come from???

An engineers handbook maybe

https://inspectapedia.com/electric/Moto ... ectionTips this might get into the ball park