Member postings for Les Jones 1

Have you considered using heat sink resistors such as these screwed to the metalwork of the machine. I think raising the temperature of the metal above dew point is better than raising the temperature of the air around it above dew point. They are available in various values and power ratings. You can use a series or parallel combination of these to suit the supply voltage you have available and the resistors you can get at a good price. You could also use a frost stat such at this to switch them on.

Les.

From yesterday (18t August.) they had lithium ion batteries and chargers. Thy call them 20 volt but they are the same as what is normally called 18 volt. I bought some of the 2 amp hour ones a few months ago when they were last on offer to use with an 18 volt drill whose nicad batteries had failed. I had to make an adapter from one of the old battery cases to fit the connector on these batteries. They work well with this drill.

Les.

Did you closely inspect the grating pattern on the glass to see if it had been damaged due to the read head rubbing against it or by swarf trapped between the glass and the read head ? Looking at your pictures it looks like the scale on the cross slide. (I call this the X axis on my DRO). If it is the cross slide then the cable to the scale will be flexed a lot so it is possible that a wire has broken in the cable. If you have an oscilloscope then look at the A & B signals where the cable plugs into the DRO. You will have to remove the cover on the connector. When the slide is moved you should see square waves (Or possibly sine waves depending on the design.) that are 90 degrees out of phase with each other. If the signals are square waves they should have a duty cycle of close to 50%. The amplitude of the A and the B channels should be about the same. (Probably about 5 volts peak to peak.

Les.

Start by setting the X and Y zeros of your DRO to the centre of the flywheel. Calculate how many degrees each segment is at the outer diameter allowing for the radius of the cutter. work out how many degrees between the sides of each spoke again taking into account the radius of the cutter. Start with the rotary table set to zero degrees. Rotate the table to half the angle of the spokes. Now take the cut swinging backwartds and forwards between that point and that point plus the angle of the segment to be removed. when you have gone all the way through add the width of a spoke angle to the end of that segment position and repeat the process until you have made the outer cut of all the segments. Set the table back to zero degrees. (I will asume the sides of the spokes will be cut in the Y direction.) MOve the table in the X direction untill the cutter lines up with the start of the first slot. You will have to adjust the Y position very slightly. Cut the side of the spoke by moving the table in the Y direction. Repeat for that side of all the spokes. Back to the first spoke move the X axis to the other side of the spoke and repeat the process cutting the other side of each spoke. When you have done all the spokes set the X axis back to zero. Position the cutter at the inner end of one of the spokes by adjusting the Y axis and rotary table angle. Cut the inner radius to the next spoke rotating the table and judging by eye when yo meet the next spoke. You can calculate these angle but remember that with parallel spokes the angle og the inner cutt will be different to the outer cut.

Les.

The speed contoller is almost certainly a PWM controller. It would be possible to design in it current (And voltage.) limiting circuitry. It could limit the PWM duty cycle if the set current of voltage limit was being exceeded. In your application the motors should share the current pretty well.
Robin, Why do you say CHEAP controllers use PWM. A linear controller for this sort of power would generate a lot of heat and waste energy. The inductance of the motor windings helps PWM to work by trying to maintain current through the winding during the off state of the PWM cycle.

Les.

The motors will only share the current equally if they are providing the same torque. (Current is about proportional to torque. ) To answer your question properly we would need FULL details of the speed controller.

Les.

You could use ignition pulses taken from the contact breaker and feed them into a frequency to voltage converter IC (Such as an LM2917) and connect the output from that to your voltmeter.

Les.

I think you will find that the 3/8" BSP thread is left handed. Would a coupler such as this solve your problem ? You could silver solder it to the UNEF part.

Les.

Does the brown wire from the run capacitor (10 uF) connect to the same terminal on the start capacitor (100 uF) as the yellow wire ? Is the wire connected to the left hand push on connector yellow and green and is it connected to earth ? Where does the blach wire go to that disappears below the left hand push on connector ? Where do the earth, live and neutral on the incomming power cable connect ? I suspect that you will have to open the motor to get directly at the ends of the windings to reverse the motor.

Les.

The main winding is connected to terminals 5 and 7 on the Duhurst switch.. The start winding with the centrifugal switch and capacitor in series with it connect to terminals 2 and 6 on the Dewhurst switch. You may need to post pictures of the connection box on your motor if the above information is not clear.

Les.

I would first check that the power from the battery holder was getting to the PCB. There could be corrosion on the battery holder contacts. ICs are normally very reliable so I think it is more likely the fault is some other component or connection.

Les.

I will have to start going back to look at the posts rather than just relying on memory. That's twice in this topic I have credited the wrong person with a comment.

Les.

Thanks Jason,

Les.

Hi Michael,
You were correct when you commented the the figure I gave (In my post on 28/7/18 at 20:05) of 5/8" bore was odd on a metric lathe. I have just measured them again and they are 16mm so the 127T gear that Ketan has show will fit my DB-10G without any modofication. Could one of the modorators please edit my post on 28/7/18 at 20:05 and add the comment "(Correction the bore is not 5/8" it is 16mm )" after where I said the bore was 5/8".

Les.

I have mounted an indexing plate on my DB-10G.

I then went a step further and coupled a rorary table to the spindle.

This is my toolpost mounted drill.

Les.

Les.

Edited By Les Jones 1 on 02/08/2018 20:14:43

Martin beat me to pointing out about not disengaging the half nut. One more point is retract the tool before reversing the lathe as the backlash in the gearing will mean that the tool will be in a different position with respect to the cut thread in the reverse direction. Also take it far enough from the start of the thread in reverse so that all the backlash is taken up in the forward direction before the tool starts to cut on the next pass.

Les.

Edited By Les Jones 1 on 30/07/2018 21:31:41

Hi Ian,
The half nut will only engage with the leadscrew when it's thread is lined up with the thread on the leadscrew. You need to gently move the half nut lever and wait until it falls into the thread. I would normally turn the lathe by hand when cutting a thread up to a shoulder. (Or use the method of running the lathe in reverse starting on the relief at the shoulder. To do this you need the tool cutting the back of the thread or at the front with the tool upside down. Someone posted a link to a youtube video showing this method recently in another thread.)

Les.

Edited By Les Jones 1 on 30/07/2018 20:48:22

Hi Martin,
I was also thinking that Ian could have the 40 and 120 tooth gears swapped over but when I worked it out I would expect that to give 127 TPI. (14.11 x 9) Also the ratio of the TPI values between gearbox position 1 (1 : 1) and gearbox position 9 (1 : 0.6) Ian quoted about 22 TPI in gearbox position 1 and about 24 or 25 TPI in gearbox position 9.
If he had 22 TPI in position 1 then I would expext about 37 TPI (22/0.6 = 36.66)
Looking more closely at the chart in Ian's second post (22:09 on 22/07/18) there seems to be a difference between the ratios between fine feed and screwcutting. There may be a control that changes the ratio or fine feed may drive the carriage via the rack and pinion rather than the leadscrew.

Les.

Can you can confirm that the 80 tooth and the 120 (Which are on the same shaft.) rotatate exactly one revolution for two revolutions of the chuck ? Can you confirm that the 40 tooth rotates three revolutions for two revolutions of the chuck ? Can you confirm that with the gearbox in position one that the leadscrew rotated exactly one revolution for one revolution of the 40 tooth on the gearbox input ? Can you confirm that with the gearbox in position nine that the leadscrew rotated exactly three revolutions for five revolution of the 40 tooth on the gearbox input ?

(I am assuming that the picture does NOT show it set up with the actual gears. I assume you have just used a picture with some other setup and marked it wit the gears that you have installed.)

Les.

The Chester DB-10G lathe I have uses modulus 1.0 gears with a 5/8" bore (Correction the bore is not 5/8" it is 16mm.) I think it is likely that the DB10 uses the same gears. If this is the case then the outside diameter of the gears = (number of teeth + 2) * 1mm. This would mean thet a 127 toothe gear had an outside diameter of 129mm. Arceurotrade sell a 63 tooth gear for the mini-lathe which is modulus 1.0 and has a 12mm bore. Ibougt one of these and bored it out to 5/8" and this give me some more screwcutting options. Chester normally have a box of assorded gears for sale at the shows. I have bought a number of different gears to give even more options. They all neaded boring out or bushing but it was worth the effort. Using the 63 tooth gear and a 25 tooth gear bought from the assorted gear box I can cut a near perfect 14 TPI thread which is actually 13.999 TPI.

Les.

Edited By JasonB on 03/08/2018 13:21:53