After reading Tom Begnal’s recent article on hybrid table saws, I have the following question:
Why does the stated horsepower differ from horsepower calculated by taking stated amps times assumed voltage of 115 and dividing by 746 watts/HP? The following table shows my calculations.
| Model<!—-><!—-> <!—-> | Stated HP<!—-><!—-> | Amps<!—-><!—-> | Assumed Volts<!—-> <!—-> | Calc HP<!—-><!—-> |
| <!—-> <!—-> | <!—-> <!—-> | <!—-> <!—-> | <!—-> <!—-> | <!—-> <!—-> |
| Crafstman 22124<!—-> <!—-> | 1.75<!—-><!—-> | 15<!—-><!—-> | 115<!—-><!—-> | 2.31<!—-><!—-> |
| Delta 36-717<!—-> <!—-> | 1.75<!—-><!—-> | 15<!—-><!—-> | 115<!—-><!—-> | 2.31<!—-><!—-> |
| Dewalt DW746X<!—-> <!—-> | 1.75<!—-><!—-> | 15<!—-><!—-> | 115<!—-><!—-> | 2.31<!—-><!—-> |
| General 50-220C<!—-> <!—-> | 2<!—-><!—-> | 13<!—-><!—-> | 115<!—-><!—-> | 2.01<!—-><!—-> |
| Grizzly GO478<!—-> <!—-> | 2<!—-><!—-> | 20<!—-><!—-> | 115<!—-><!—-> | 3.08<!—-><!—-> |
| Hitachi C10LA<!—-> <!—-> | 1.5<!—-><!—-> | 15<!—-><!—-> | 115<!—-><!—-> | 2.31<!—-><!—-> |
| Jet JPS*10TS<!—-> <!—-> | 1.75<!—-><!—-> | 17.5<!—-><!—-> | 115<!—-><!—-> | 2.70<!—-><!—-> |
| Shop Fox W1748<!—-> <!—-> | 2<!—-><!—-> | 20<!—-><!—-> | 115<!—-><!—-> | 3.08<!—-><!—-> |
| Steel City 35601<!—-> <!—-> | 1.75<!—-><!—-> | 15<!—-><!—-> | 115<!—-><!—-> | 2.31<!—-><!—-> |
| WoodTek 148-271<!—-> <!—-> | 2<!—-><!—-> | 18<!—-><!—-> | 115<!—-><!—-> | 2.78<!—-><!—-> |
HP = (Voltage x Amperage) / 745.6
Replies
Actually, the applicable formula is voltage x amps x efficiency divided by 746. The 746 watts per horsepower is for a perfect motor. Real motors are less than perfect, and the Asian motors placed on inexpensive power tools are considerably less than perfect. Some of that theoretical power, probably around 20%, is lost as heat, noise and stray magnetic fields.
On top of that, I am highly suspicious of the labeling on the import motors. I'm pretty sure that the values given on the nameplate are, at best, only an approximation of the real current draw. Fairly often at FWW we would test two nearly identical tools with identically rated motors and they wouldn't perform the same at all, one tool having considerably more power.
John White
Thanks for the quick response.
Stephen in New Mexico
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