. American engineer and railroad journal . nal. We wish to have a range of speed upon this tool offrom 8 to 240 rev. per mln. The class of work to be done onthis latho will not be very heavy and it was decided to fit Itup for a capacity of taking a /4-in. cut with a 1-32-ln. feed onhard steel at a cutting speed of 50 ft. per minute. The maximum horse-power required to do this work =: 1-32X 1/4 X 50 X 12 X 1 X .7 = 3A- (See formula on page 125 of thepreceding article of this series.) Since the work is of a variablediameter, it will require 3V4, or say 3 horse-power, throughoutthe entire speed r
. American engineer and railroad journal . nal. We wish to have a range of speed upon this tool offrom 8 to 240 rev. per mln. The class of work to be done onthis latho will not be very heavy and it was decided to fit Itup for a capacity of taking a /4-in. cut with a 1-32-ln. feed onhard steel at a cutting speed of 50 ft. per minute. The maximum horse-power required to do this work =: 1-32X 1/4 X 50 X 12 X 1 X .7 = 3A- (See formula on page 125 of thepreceding article of this series.) Since the work is of a variablediameter, it will require 3V4, or say 3 horse-power, throughoutthe entire speed range. The question which naturally presentsitself is: Can we make use of the gearing now on the lathe Inconnection with the application of the motor? The presentback gear ratio is to 1. Assume that the motor is con- 100I 90 L • SO t • 70? I 60 a. : 5[40 i 30 1^ ao 3 I 10 _fu// line shOius autrage speed• Dotted lint shows powerSolid oirciea ahow fundamental or uoltage circles show derived orinterinaJute afieeds. ^-y-. OS 03 1 10 11 6 12 7 J-S. 3 4 5 6 7 8 Controller poaltiont, forward2 3 4 Controller uositions, reverie FIG. 2.—DLAGRAU SHOWLNQ RELATION OF THE SPEED AND POWKB OF A-MOTOB USING THE M. A. 12 (aBMATDBE RESISTANCE) CONTROLLER. nected so as to drive the main spindle at 240 revolutions perminute. Now, if the back gear is thrown in, the spindle speedwill be 240 -=- =; revolutions per minute. In order tofill in the gap between 25 and 240 revolutions per minute themotor would be required to have a speed range of about 1. Referring to Fig. 1, it will be found that when the motor is running at per cent. Q« its maximum speed, or the lowest speed, it will furnish about 20 per cent, of its ratedpower. Therefore, in order to furnish 3 horse-power through-out the range, a 15 horse-power motor would have to be motor of this size would be very large and bulky for a and would not be very efficient at the lower speeds.
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