Calculating Bathmates vacuum?
First time poster, kinda feel this isn´t the right section for this topic but will do.
So, got annoyed with the whole “how´s the Hg when you pump” related comments to end arguments against this machine so wanted to calculate exactly how much we are using. And I´d love some comments from people that are smarter than me and even better if with some know how to physics of the subject.
Anyway so far we can determine that due to the gaiters spring the more it´s compressed the more force it uses to try to return to it´s neutral position. Simple, pump it till it stays down 1cm or max, the difference is simple enough to notice, while in it. Due to Newtons third law, the force we apply on something is opposite at the other end, we can determine the force applied when the spring is pressed down X amount. Take a regular person weight scale, press your BM against it and see the weight you are using, that is the weight the spring is pressing back up, once the spring is fully pressed any additional weight will not be the springs doing but go through directly to the scale. So basicly once the spring no longer compresses stop and that´s the max force the machine can ever create.
Now KG is easy to make into N just weight*gravity, so a*9.8= b N
Simplest solution I could think of is F=AP(force, area, pressure) we know the force, we know the area, we can solve the pressure. Now before I get butchered by a physic master, I´m almost certain this gives you a sort of “absolute” measure, without taking into account how much water expands and perhaps in what temperatures. As far as I know not much, anyway lets continue.
By my own testings, with a X40 BM the closer the max compression comes the more the weight goes up so it´s hard to determine for 100% but seems that 30KG is the max, after that it´s hard to tell, now using the previous equations 30kg is 11inhg which seems inline with BM claims though I do remember it being said for the old models not the X, and as well a memory that it´s the max the cap will hold, so not really the max the unit can do. So by this logic if you use 30kg of force to press that thing on your groin(it´s alot) to reach the very bottom you AND pump it till the pump stays at the very bottom you´ve reached 11Hg. Lets face it pumping this thing to truly max so that it stays there is hard to impossible few MM higher and you lose some 5kg already.
I say again how much you press down on it doesn´t matter what matters is where does it stop when you release it.
So that stuff above is what I´d like some more professional opinions on. If that is ok, then using the same logic just reverse engineered I can determine the following for my unit. 8kg=3Hg, 11kg=4Hg, 13.6kg=5Hg So now I have to measure how much the spring is pressed to reach those weights. For example ruler against the side measuring from the bottom of the unit to the first number and presto I have invented a -not so accurate- pressure gauge.
I feel it can´t be that easy, would have been done already. But my logic tells me that this is the highest possible value if water expands it goes lower, change in temperature is admittedly a factor these are on inHg 60F. Also the length of the tube, but we are in so small amounts here it can´t be that much, can it? And another that pops to mind is that the BM is narrower on top, might be a factor.
Self critic, the diameter of the surface plays a huge role in this, if the force remains the same but the diameter is less it goes up quite a bit, as an example if I calculate only the inside diameter(the area with water not the edges included) I´d get 17inHg. Must say though that if they reported the hercules model being 11Hg these are supposed to be 30% more so at max this should be 14Hg by that logic, *shrug*. So like said in the very beginning these are just my thoughts and calculations and would love some feedback especially from people that know stuff like this. The main reason I´m “excited” about these results is due to the 11Hg which is in line with what the corporation has stated. I admit these can be totally wrong but through mistakes we can learn.
Oh and then the calculations,
Radius of the outer edge of the bottom of 40X is 5cm.
P=F/A N/(Pi(radius)^2), 294/(3.1416*(0.05m)^2)=37 452Pa =11.1inHG 60F