SpudFiles

for a coaxial, how far back of the chamber should a piston move? Also, how big should equilisation(sp?) hooles be?

1/4 of the barrels I-D, is how far it should only need to move back.

EQ holes are only needed if your piston is pretty much "air tight" to start i would go without them/one. then if filling is to slow maybe put a really really lil hole for EQ. like im talkin >.5mm.

Right...but then the air from behing the piston wouldn't fill the chamber if it was air tight...

You only need an equalisation hole if your piston is 100% airtight (ie you're using 0-rings or a syringe seal or similar), otherwise it's unecessary.

And any piston movement more than 1/4 of the caliber is a SIN AGAINST EFFICIENCY and will be mocked with great vigour

Hi jack... (etc)..

You are getting be curious; why is it a sin to have more piston travel than d/4?

I can imagine it wouldn't do much good, but where's the bad? Increased pilot volume?

Regards
Soren

That and the piston accelerates when you fire, so the further it moves the greater the momentum it has.

To me a 1/4 of the diameter seems abit on the small side......

Michael

Gepard wrote:
To me a 1/4 of the diameter seems abit on the small side......

Michael

Yeah I was wondering about that... Does anyone know where that information came from? I'm sure it's correct but it would really help to know exactly why it is true.

In fact a quick bit of maths suggest that d(3/4) is more appropriate....

Let's say you have a pipe with an inside diameter of 90mm. That has a cross sectional area of 6361mm^2. Now to get a rectangle of that area (assuming that the inside of a T is rectangular) you need to have a rectangle of size 6361/90 = 70mm

So the piston should slide back 70mm not the 1/4*90=22.5mm that 1/4 gives....

Or have I screwed up there? Probably since it's nearly 2am here.

I realise I've rounded probably excessively but still that is a massive difference.

Michael

Hi Gepard

Looks OK, except that you rounded to 70mm, not 71.

I agree with you most of the way - except maybe: The area to use as a standard flow area is not necessarily the valve seat area. If the barrel area is smaller - eg a 75 mm pipe, which had ID = 69 mm, then the same open-valve-flow-area could be reached by:

(69/2)^2 * pi / 90 = 42 mm.

For me, the conclusion it: Make pistons light, and give them a nice, long travel. That with the light pistons is easy to say for me, after I found this series of metric parts where the endcaps seem to be designed to serve as pistons in the tees .

Regards
Soren

I don't feel like doing the math (it's my last workday in Spain so I got completely plastered last night ) but if you had to take a tube that is the same diameter as the inside of your barrel, and cut off a section that was 1/4th the calibre long, the surface area of this tube would be equal to the area of your barrel.

So basically - small, lightweight pistons with minimal travel are ideal.

The advantage of keeping pilot volume small, aside from faster opening time, is that you don't need a high flow valve to actuate it - so far all my pistons up to 1" have been actuated by the same schrader I use for filling.

dongfang wrote:<snip>
(69/2)^2 * pi / 90 = 42 mm.<snip>

Where does the 90 come into it?

I wouldn't have said you want "nice, long travel" for all the reason listed you just need it large enough to not restrict the air flow.

Jackssmirkingrevenge,

You don't need to do the maths as Dongfang and I both did it and found that 1/4 of the diameter in length isn't equal to the actual tube's surface area.

Does that mean then that people have been building underpowered piston valves then?

Let me put it into inches to invite some of you Americans into the discussion

Say you have a pipe of inside diameter 1".
Surface area is therefore: pi*0.5^2=0.785 inches squared
1/4 of the Diameter (ie Piston travel): (1/4)*1"=0.25"
Cross section(piston travel x internal diameter)= 0.25*1=0.25 inches squared

Michael

let's assume a 1 inch barrel then (I'm not American btw ;p)

area of the barrel (the "flow" we wish to achieve) is pi*r² which in this case is 3.142*0.5*0.5 which equals 0.7855 in ²

for 1/4 of the caliber of travel, you get an available area of 2*pi*r*h, in this case 2*3.142*0.5*0.25, which equals 0.7855 in ²

If I'm wrong, I blame the hangover.

If I'm right, QED

Why's the second equation times two?

Michael

Gepard wrote:Why's the second equation times two?

because the Surface area of a cylinder (ignoring the ends in this case as we only want the area of the side of our virtual "cylinder") is the circumference multiplied by the height, 2 * ¶ * r * h

Hmmmm makes sense but I still that a bit of extra travel would be beneficial....

Michael