Gas Strut for Mill

I have to replace the gas strut on my Toolco on VM32 mill (similar to Warco WM18). Good time to do it as the mill is in bits midway through a CNC conversion. The original gas strut has competely failed and there is no info on the cylinder to tell you the force in newtons to compress it.

I can tell you the weight of the head including motor and saddle is 55kg.

Elsewhere i have read that the strut should be equal to the weight of the head and so i bought a 550 Nm strut but i don't think thats right. I can't even start to compress it. Also the strut will require more force as it compresses the nitrogen. The strut only needs to stop the head falling on the ballscrew. With the acme screw it always stayed put.

The strut is fitted perpendicular inside the column

Your opinions or advice is most welcome

Brian

Edited By Brian Rutherford on 20/05/2020 16:18:41

I bought mine from here SGS engineering. Some information on sizing, fitting and use. Mine have been in my machine for a couple of years with no issue. Usual no connection with company just good service.

A picture in my album of assembly used. I think it balances about 15kg on a Sieg SX2P quite nicely, just very slight lift.

Interstingly my KX3 just has a big coil spring rather than a strut and the X3 nothing

Thanks for the replies. Howard I bought the 550nm fom SGS but will be exchanging for a lower value one. I know exactly how it fits. What i need to know is what value. If you are using a 15kg one on a Sieg sx2. That equates to approx 150 Nm. I would hazard a guess at the head on yours would weigh about 40 kgs therefore possibly a 200 Nm one might be adequate.

Howard just looked at your album. The strut on mine is factory fitted inside the column. I have emailed toolco for the info but had no reply. Their phone  number is discontinued

I see you are using 2 struts so i would need to double the value as mine is a single strut

Edited By Brian Rutherford on 20/05/2020 17:36:18

When i installed my wm18 mill i removed the gas strut and couldn’t begin to compress it

seems to work ok on the mill

sorry not much help as to what rating you need

Just being pedantic perhaps, but Nm is torque or work, not force. Are gas struts rated by some odd units?

That SGS link is useful, it is going into my favourites in the browser. The knee of the Tom Senior could do with a bit of assistance, especially when heavy kit is on the bed. I will be checking the dimensions and will have a look to see if they have a suitable strut.

The SGS site rates struts in Newtons, not newton metres.

Edited By old mart on 20/05/2020 18:19:28

Ian, that helps a little. The strut on mine failed 12 months ago and it made some difference to raising the head. I want to replace it now for two reasons. One too make it easier on the motor and two, so the head wont drop when the motors arenot under power. Never happened with the acme screw, but will with the ballscrews.

The fact you couldn't compress the strut on your mill makes helps

The strut assembly fits to the top of the column with the struts down inside. A cable fitted to the head at one end goes round the pulleys and fitted to the underside of the plate. So the travel was halved meaning twice the load. This arrangement was because the column height didn’t allow enough length for the strut stroke and also gives a straight load on the strut. In all it has worked well.

I recently replaced one in the column of my drill, you won't be able to push it by hand, they kinda lock out at the end of travel, once it has been compressed an inch or so it becomes much easier to compress.

Think about when you're closing the boot on your car, the initial force to break it free (while it's fully open) as the struts are at the end of their travel, then the force to close it becomes much less.

Thanks Pete, Seems likely i have the right strut. Not sure if i will be able to fit it as i think i need to compress it an inch to get it in. I'll have another look at it in the morning.

About a year ago I ordered a 6mm strut at £27 for my sx2p but it was out of stock.

I noticed that they have now reduced the price of the same strut by £6

Unfortunately it is still out of stock!!!!

Brian, the head of my drill weighed about 40kg, I first tried a 400n strut, but it was too strong, I then tried 350n strut, it was about right, but maybe still a little too strong, it's difficult too calculate exactly, if I was doing it again, I'd buy an adjustable gas strut, if my memory serves me right, they were 20 something quid, that way you can fine tune it.

The benefits of hindsight.

But like I say, don't worry too much about the initial compression, even really low weight rating struts can be difficult to compress without leverage.

Thanks Pete

I have sent the strut back and am replacing with a 400 newton one . Also 50mm shorter, so should be able to fit ok

Yes, sometimes! The figure is torque on the assumption the strut is for use in a known hinged arrangement like a car hatchback. I guess it's convenient buying spares for particular vehicles or something. But really they should be in Newtons. From this website:

Maths make my head hurt, but you're good at this stuff Duncan. Assuming the strut is vertical, and the head is 'x' kilograms, what formula would give the strut in Newtons?  (I think it's the above with d = 0)

Buying a strut for a mill, I would always go for one calibrated in Newtons rather than Nm, mainly because I don't have a feel for the difference! Working in kg for struts calibrated in N is inside my comfort zone and 1kg being close enough on earth o 10N for most practical purposes keeps the arithmetic simple. (And I need all the help I can get.)

Dave

Edited By SillyOldDuffer on 22/05/2020 10:36:19

SGS will also regas your struts if suitable.

I don't know where the 1.06 factor comes from, but otherwise the sum is correct. The cosine of the angle twixt strut and vertical (call it alpha) is (c/d)

the vertical force required is F*(L/d), and the vertical component of the cylinder force is F1*COS(alpha), so for equilibrium

F1*c/d = F*L/d.

multiply both sides by d/c gives F1 = F*L/c. Perhaps the 1.06 is to make it stay up, not be just in balance. If the strut is vertical, c = d.

Still doesn't involve Nm, which may well have been an imsprint