Okay ... I've BEEN to wikipedia AND Googled for about 20 minutes

if you recall that one atmosphere is 14.7 PSI, this chart will help:

and if you further reacall that one atmosphere is 30 inches of mercury (aprox), then it should all be obvious. You don't need a vacuum less than .07 bar or so, that's 70 milibar, or 7X10E5 greater than the value you have below. You REALLY REALLY REALLY do NOT NOT NOT want a high vaccuum pump. you may find my article on vacuum pumps and chucks helpful - you can download it from my web page click on vacuum pumps) or you can download it from the tips section of - it's the same article both places. bill (not in Detroit)

Not sure I understand you: a hectopascal IS a millibar. Shipping forecasts and reports here (in France) now use the hectopascal as standard for all atmospheric pressures. For medical purposes (the only area where it is still used in Europe, apparently), 1 mm mercury = 133.322 pascal, or 1.33322 hectopascal.

As 1"=25.4 mm, my calculation makes that 1" mercury = 33.864 hectopascal.

Unless I've screwed up somewhere...

only one p in my real address / un seul p dans ma véritable adresse

A pressure of 1 Atmosphere = 30" mercury = 760mm mercury = 15lbs/sq" = 760 Torr = 32ft of water = 1 bar = 1000mbar = 1013 hectopascals. (all approx.)

Assuming that the question relates to vacuum chucks.

A metric/imperial conversion shows that for every 67mbar below 1000mbar you will have 1lb/sq" holding the bowl in place.

If you get down to 67mbar (1000-14x67mbar) you will have 14lb/sq" holding the bowl in place. Reducing the pressure below 67mbar is an attempt to gain the extra 1lb/sq" to hold the bowl in place. I would suggest that if the anyone struggling for the last 1lb/sq" is living life on the edge.

I'm sure Bill Noble's advice would be to make sure the pump has the throughput to the maintain the target vacuum whilst handling leaks of the seals and porosity of the wood and release of water from the wood at reduced pressure.

I hope this helps.

BillR

Millibar. One thousandth, right? Nope. BTW, 29.92" or 760 mm is assumed standard sea level pressure in Hg . Mystery is why there are 1014 one thousandths of a bar in a standard atmosphere.

You might find helpful.

--

--John to email, dial "usenet" and validate (was jclarke at eye bee em dot net)

Thanks, Dan. I couldn't find the math to make the conversion and without some simple 'rule of thumb' to use in comparisons, I couldn't tell if what I was looking at was in any respect appropriate - or not.

I've been to Bill Nobles site a number of times, but he uses in/mm and most of the other sites a casual Googling turned up used other units and no one was providing any sort of a Rosetta stone.

Bill

Thanks, Bill. I don't think I've needed to know either of those numbers more than once or twice since graduating from High school in 1970. And possibly stretching all the way back to Jr. High.

I see various units offered in the catalogs, your offerings, and the other offerings on the internet and I was trying to puzzle out just what sort of unit would have more than enough suction without sucking all the money out of my pocket or requiring an underground bunker for noise abatement.

Out of curiosity, how far away from what I need would a good shop vac be? I mostly want to finish turn the bottoms of bowls in the 6-10" range with flat edges that have already been sealed.

Hmmm ... maybe I could put a barometer in a wooden box with a plexi face and test for myself. That should be good enough to give me a ballpark feel for what's going on.

Bill

I just tried google with the following search parameters..

convert millibars to hg

The second URL in the results was

amd it had the following notes To convert inches of mercury to millibars, multiply the inches value by 33.8637526 To convert millibars to inches of mercury, multiply the millibar value by 0.0295301.

Hi Bill

Bill I'll try to make things a little simpler, that works just fine for everyday approximation, like a circle is about 3/4 the area of a square, some will cringe but yes thats good enough for what we need in most cases.

Also with vacuum, take all air away and you would have 15 # pressure per inch square

The 15 # is the same as 30" mercury, or 75 cm mercury and that is also

1 bar.

To get 1# of pressure per square inch you'd need 2" mercury or 5 cm of mercury.

So now you have a let's say 10" Diameter bowl, that would be 10x10=100, and we take 3/4 of that for the area of a circle and we have 75 square inches.

Now if you had a vacuum of just 4" or 10 cm of mercury and no leakage you'd have 150 pounds holding your bowl down, that should be lots, but.............

Next you have a 6"D bowl, that would be 6x6=36, and 3/4 of that is 27" square with no leakage and 4" or 10 cm of mercury vacuum and 54# holding it down should also be enough to clean up the bottom

Now we take a 3"D bowl and 3x3=9, 3/4 of that gives us 6.75" square, no leakage, and 4" mercury vacuum 10 cm or 100 mm same as before, and now we'd have only 13# holding it down, I would suggest that's not enough.

If you now increase the vacuum to let's say 20" of mercury, you'd have

10# of pressure per square inch, and on that 3"D bowl the hold down would be 67.5# and that would be lots IMO.

So do you need all that high vacuum ??, I would say no, but you do need enough volume though, as there is always leakage and if you don't have the volume you'd never get any vacuum.

A good shopvac will give you enough vacuum for a larger bowl, however you better keep in mind that the vacuum motor needs air to flow by to cool it, and if you have very little air going past, like with using it as a vacuum holding pump your shopvac's motor will burn out, period.

Hope this helps you and anybody else that has some questions about all of this.

Have fun and take care Leo Van Der Loo

Bill > How can I correlate mbars, as in :

a really good shop vac might draw 5 inches of mercury. The table in my article will show you the holding force, but it's basically 1/6 atmosphere, right? or about 2.5 PSI of force - if you figure you need 100 pounds of force to hold your object (that's just a number pulled out of nowhere, but it is probably plausible), then your shop vac would be OK with anything whose diameter (remember area = pi times radius squared) is greater than 7 inches or so

try this table (view it in a fixed pitch font) - hopefully I made no errors in the spreadsheet

Force for indicated diameter in inches in mercury psi 1 2 4 6 8 10 1 0.49 0.3848451 1.53938 6.157522 13.85442 24.63009 38.48451 2 0.98 0.7696902 3.078761 12.31504 27.70885 49.26017 76.96902 3 1.47 1.1545353 4.618141 18.47256 41.56327 73.89026 115.4535 4 1.96 1.5393804 6.157522 24.63009 55.41769 98.52035 153.938 5 2.45 1.9242255 7.696902 30.78761 69.27212 123.1504 192.4226 6 2.94 2.3090706 9.236282 36.94513 83.12654 147.7805 230.9071 7 3.43 2.6939157 10.77566 43.10265 96.98097 172.4106 269.3916 8 3.92 3.078760801 12.31504 49.26017 110.8354 197.0407 307.8761 9 4.41 3.463605901 13.85442 55.41769 124.6898 221.6708 346.3606 10 4.9 3.848451001 15.3938 61.57522 138.5442 246.3009 384.8451 11 5.39 4.233296101 16.93318 67.73274 152.3987 270.931 423.3296 12 5.88 4.618141201 18.47256 73.89026 166.2531 295.561 461.8141 13 6.37 5.002986301 20.01195 80.04778 180.1075 320.1911 500.2986 14 6.86 5.387831401 21.55133 86.2053 193.9619 344.8212 538.7831 15 7.35 5.772676501 23.09071 92.36282 207.8164 369.4513 577.2677 16 7.84 6.157521601 24.63009 98.52035 221.6708 394.0814 615.7522 17 8.33 6.542366701 26.16947 104.6779 235.5252 418.7115 654.2367 18 8.82 6.927211801 27.70885 110.8354 249.3796 443.3416 692.7212 19 9.31 7.312056901 29.24823 116.9929 263.234 467.9716 731.2057 20 9.8 7.696902001 30.78761 123.1504 277.0885 492.6017 769.6902 21 10.29 8.081747101 32.32699 129.308 290.9429 517.2318 808.1747 22 10.78 8.466592201 33.86637 135.4655 304.7973 541.8619 846.6592 23 11.27 8.851437301 35.40575 141.623 318.6517 566.492 885.1437 24 11.76 9.236282402 36.94513 147.7805 332.5062 591.1221 923.6282 25 12.25 9.621127502 38.48451 153.938 346.3606 615.7522 962.1128 26 12.74 10.0059726 40.02389 160.0956 360.215 640.3822 1000.597 27 13.23 10.3908177 41.56327 166.2531 374.0694 665.0123 1039.082 28 13.72 10.7756628 43.10265 172.4106 387.9239 689.6424 1077.566 29 14.21 11.1605079 44.64203 178.5681 401.7783 714.2725 1116.051 30 14.7 11.545353 46.18141 184.7256 415.6327 738.9026 1154.535

"Bill in Detroit" wrote in message news: snipped-for-privacy@corp.supernews.com...

Bill, not my day.

Are the numbers 1-30 along the left hand side just row numbers? If so, I think I could reconstruct the spreadsheet. If not, could you supply the missing labels? Read R -> L as I have the least clutter in the far right column.

TIA

Bill

well, I didn't save the spread sheet - but here's what I did

left most column - inches of mercury next colum, PSI (due to air pressure pushing on the item) - = inches of mercury * 30/14.7 next columns, force due to pressure - numbers across the top are diameter, calculated by =(diameter/2)^2*psi

I believe I put a similar table near the front of my treatise on this subject (see

or