I guess the formula i'm looking for would be a function of the distance between the lever pivot and the handle, the distance between the pivot and where the piston is attached to the lever, and the radius of the cylinder.
Is there any engineers up to the task?
Calculating the force on the level required to reach 9 bars?
- HB
- Admin
Steve's empirically calculated the results with a cheap bathroom scale and a brew pressure gauge. See Elektra/Pavoni side by side and Olympia Cremina 2002: The evolution of design, excerpted below:
To make the second chart, he read off the numbers to a tape recorder as he pulled the shot, and then later transcribed it.srobinson wrote:What we rigged up was an elbow pressure gauge on the bottom of the Olympia portafilter. My wife currently has my digital camera at her parent's house, but I will post some pics of the setup. With a real pressure gauge I hypothesized that I could be a bit more exact on how much downward pressure would be needed to record specific bar pressure on the gauge. Utilizing my handy bathroom scale, that became famous in my Elektra/Pavoni comparison, I was able to plot the following curve:
With this data as my reference curve, I did a rough plot of the pressures that I saw during a good shot at 2 second intervals. To help understand the line, the shot had a 10 second pre-infusion, followed by a first pull...and then you will see the line drop to 0 as I did a second pull until blonding. Take a look:
A couple observations from this data. First, there is very little reference material out there that helps me validate these curves. There have been several posts stating that it takes roughly 40lbs to pull 9 bars, so I was pleased to validate that point. I think the pattern does give you some insight on the unique nature of lever machines. At the start of the pull, infusion is under your control and you can hit the puck immediately with full pressure. There is no ramp-up. Also with these machines, there is variable pressure during the pull. I was using a pulling style of letting the machine do the work and tried to stay with its resistance. The final observation is that with a full manual, I believe that you are able to start at a higher pressure and maintain an overall higher pressure than with an spring loaded machine.
Dan Kehn
The problem is that i expect that all lever machines are slightly different when it comes to the pressure needed on the lever...and i don't have the equipment to do any measurements myself.
However, i believe this is correct, in a static condition where the machine is completely choked:
F = (p * pi * r² * La)/Lb
"p" is the brew pressure in bars (900,000)
"pi" is pi (3.14...)
"r" is the inner radius of the cylinder in meters
"La" is the distance between the two screws on the lever (from the screw that attaches the lever to the group to the screw that attaches the lever to the piston)
"Lb" is the distance from the screw that attaches the lever to the group to the point at the handle where you apply your pressure.
"F" is the force that you use to pull the lever, in newtons. Multiply that with 0.22 to get whacky non-standard units (pounds;))
Anybody with a Cremina who would care to take the measurements so that we can compare the formula to the diagram?
However, i believe this is correct, in a static condition where the machine is completely choked:
F = (p * pi * r² * La)/Lb
"p" is the brew pressure in bars (900,000)
"pi" is pi (3.14...)
"r" is the inner radius of the cylinder in meters
"La" is the distance between the two screws on the lever (from the screw that attaches the lever to the group to the screw that attaches the lever to the piston)
"Lb" is the distance from the screw that attaches the lever to the group to the point at the handle where you apply your pressure.
"F" is the force that you use to pull the lever, in newtons. Multiply that with 0.22 to get whacky non-standard units (pounds;))
Anybody with a Cremina who would care to take the measurements so that we can compare the formula to the diagram?
- hbuchtel
Nice!Staffan wrote: F = (p * pi * r² * La)/Lb
I really don't have a head for this type of thing, can you tell me if the same formula would work for a two-handled lever machine? (Presso)
Thanks, Henry
LMWDP #53
I believe that it would work, and the force that you would get is the combined force needed on both levers - like it would read if you put the machine on a scale and pulled a shot. However, i guess it would be difficult to measure towards the end of the pull, since you'd be pushing the levers together rather than down on the scale.hbuchtel wrote: I really don't have a head for this type of thing, can you tell me if the same formula would work for a two-handled lever machine? (Presso)
Thanks, Henry
If you have a presso suffering from this problem, then we'd need to get into more complex formulas, since i'm not counting with things like air in the top of the cylinder...
- another_jim
- Team HB
There's a much simpler method, measure the distance the lever handle travels, and the volume of water dispensed.Staffan wrote:The problem is that i expect that all lever machines are slightly different when it comes to the pressure needed on the lever...and i don't have the equipment to do any measurements myself.
However, i believe this is correct, in a static condition where the machine is completely choked:
F = (p * pi * r² * La)/Lb
"p" is the brew pressure in bars (900,000)
"pi" is pi (3.14...)
"r" is the inner radius of the cylinder in meters
"La" is the distance between the two screws on the lever (from the screw that attaches the lever to the group to the screw that attaches the lever to the piston)
"Lb" is the distance from the screw that attaches the lever to the group to the point at the handle where you apply your pressure.
"F" is the force that you use to pull the lever, in newtons. Multiply that with 0.22 to get whacky non-standard units (pounds;))
Anybody with a Cremina who would care to take the measurements so that we can compare the formula to the diagram?
Say, for simplicity, 45 mL water, and 22.5 cm of travel, then the "tube" of water being pushed on the lever side of an imaginary hydraulic pump is 2 cm-square. A bar is roughly 1 Kg per cm-square, so you'll need 2*9 or 18 Kg.
Jim Schulman
- hbuchtel
Could you make a version of the formula that used the vertical travel of one's hands vs the piston rather then the two sections of the lever?Staffan wrote:I believe that it would work, and the force that you would get is the combined force needed on both levers - like it would read if you put the machine on a scale and pulled a shot. However, i guess it would be difficult to measure towards the end of the pull, since you'd be pushing the levers together rather than down on the scale.
In my case I don't, but it brings up a question, are you saying water doesn't compress enough that you need to take it into consideration?Staffan wrote:If you have a presso suffering from this problem, then we'd need to get into more complex formulas, since i'm not counting with things like air in the top of the cylinder...
Thanks for bringing up this question, I don't have the hardware that Steve used (in Dan's quote above) so what you are proposing seems to be a good way to work out the answer.
Henry
- hbuchtel
uh-oh . . . would you mind going into a bit more detail?another_jim wrote:There's a much simpler method, measure the distance the lever handle travels, and the volume of water dispensed.
Say, for simplicity, 45 mL water, and 22.5 cm of travel, then the "tube" of water being pushed on the lever side of an imaginary hydraulic pump is 2 cm-square. A bar is roughly 1 Kg per cm-square, so you'll need 2*9 or 18 Kg.
Henry
- hbuchtel
Steffan wrote:F = (p * pi * r² * La)/Lb
This is what I love about espresso! Not just the aesthetic and functional beauty of the machines, the thrill of caffeine, the physical barista skills, the exploration of taste, the friends and community, now also the chance to re-learn "ancient" formulas that I never thought I would use!another_jim wrote:There's a much simpler method, measure the distance the lever handle travels, and the volume of water dispensed.
Say, for simplicity, 45 mL water, and 22.5 cm of travel, then the "tube" of water being pushed on the lever side of an imaginary hydraulic pump is 2 cm-square. A bar is roughly 1 Kg per cm-square, so you'll need 2*9 or 18 Kg.
Henry
- another_jim
- Team HB
Imagine you are using a hydraulic pump to push the water through the puck. You'd be pressing down a long distance, at less force, to create 9 bar over a narrower crosssection of water. So if the lever travels 22.5 cm, what is the crossection of the tube to get 45mL of water? 2cm-square. How much pressure is required to get 9 bar on this? 18 KG. The mechanical advantage is being provided by a lever, rather than a U or flare shaped tube of water, with one side having a small area, the other large, however, the same basic physic must apply. So it's easier to think about pushing a narrow column of water a long way rather than worrying about piston and puck sizeshbuchtel wrote:uh-oh . . . would you mind going into a bit more detail?
Henry
Jim Schulman