Since my Peppina is now temporarily out of order because i can't find my tube of Dow 111, I spent time with her today by trying to figure out how much pressure she actually develops. I've found some tidbits like "i guess around 6 bar" on the internets, but never any photographs of pressure gauges attached to her or definitive statements like "i tested it in Geneva by stacking standard kilograms on her spring and she delivers 7.43 bar max". So i thought i could at least give it a shot.
Since i don't have a pressure gauge i'm doing this theoretically, bearing in mind that pressure is force divided by surface area. The surface area is easily found. The diameter of her piston is 50mm, the radius of the circle is 2.5cm, the square of that is 6.25, times pi gives ~19.6cm^2.
Calculating the maximum force exerted by the spring was a bit harder for me to do. I don't have pressure meters that go all the way to ~200kg, but i do have a bathroom scale.
So, i put the bottom half (without kettle or group) of La Peppina on the scale and pulled down the lever. Her bottom weighs 1.7kg, and when the lever is fully down, the scale read 10.0kg. So, the force i put down on the lever was 8.3kgf.
On this diagram, its easy to see how the spring is connected to the lever

On that pic, the lever is up, and the connecting piece has an angle of ~30 degrees upwards. When the spring is fully depressed, it has an angle of 45 degrees downwards. The distance between the spring connecting rod and lever rod is 1.5cm. But at an angle of 45 degrees, the effective horizontal distance is 1.5/cos(45) = 1.1 cm. The length of the lever from the tip to the base is 30 cm (no curvature). 30/1.1 is 27.3. So the force exerted by the lever on the spring is 27.3 times as high as the force at the tip of the lever. That gives the force at the spring at 226 kgf. Which is a lot. Divided by the surface area of 19.6 cm^2 gives 11.6 bar at the beginning of the pull.
A lot more than most people thought it did. Still, i can't see any obvious flaws in my math. So it has to be somewhere in the vicinity of true.
Still don't get why La Peppina doesn't give a lot of crema.

/tries to hang with the cool kids by trying science

The bathroom scale method is too crude.

http://www.efunda.com/DesignStandards/s ... signer.cfm

They can do a nice job with the correct roast and origin. After a rebuild, this was about a year ago and with a Yergh.

Oh I do love this math and theroy. Please go with this farther.

timo888 wrote:The bathroom scale method is too crude.

http://www.efunda.com/DesignStandards/s ... signer.cfm

Cool calculator. Sadly, the manual of the Peppina doesn't state the material properties of Peppina's spring.
I fully realise that the bathroom scale is a very crude measuring device. But it is the best i have. But i still feel it has relevance, because as far as i've found, i'm the first to actually try to measure the pressure developed by La Peppina. But yes, it wouldn't surprise me if i'm a bar off.

BlueCold wrote:Sadly, the manual of the Peppina doesn't state the material properties of Peppina's spring. ... But yes, it wouldn't surprise me if i'm a bar off.

Believe me, you're not getting nearly 12 bar at the basket on the Peppina. Try cutting your estimate nearly in half. Is there any material that would yield the force required to reach 12 bar? You can count the coils and take the necessary measurements.

Pressing the lever down with the machine on a bathroom scale is hardly a "measurement". Much of your arm's force is being translated through the fulcrum to the base, not to the spring. This force would have to be subtracted before you divide by surface area.

Bluecold wrote:I've found some tidbits like "i guess around 6 bar" on the internets, but never any photographs of pressure gauges attached to her. . .

Here's one on Lino's

Hmm. That would mean my calculations are twice too high. Something, somewhere, went wrong. A bathroom scale isn't really accurate, but giving a reading twice too high seems too much. It's a digital thing. Piece of crap because it keeps turning off to save power, but it should be more accurate than one with a simple spring and a dial, shouldn't it?

As I said, it's not a problem with the scale, it's a problem with the method.

When you press down on the lever, a good deal of the force is NOT devoted to compressing the spring, but is instead translated down the axis of the lever arm and via the fulcrum to the base, and thence out to the scale.
In fact, if you press perfectly parallel to the axis of the lever arm, NONE of the force reaches the spring, and the lever arm won't move, but the scale will register your weight. Now, when truly trying to compress the spring, you're going to try to press perpendicular to the axis of the lever arm. But to the extent that you are not pressing perfectly perpendicular, you're wasting some effort, which gets translated to the base, not to the spring. This "wasted" force not applied to the spring (which is not accurately calculable using your method) must be subtracted from the total force applied in order to arrive at the force that was spent on compressing the spring.

But see how much weight you can apply to the lever arm without making it move (axis-parallel force). Note that weight, and subtract it from your earlier results. Then divide. This will yield an estimate with some attempt to reduce the total force by the amount of wasted effort.

I watched the scale when the lever was horizontal
The scale only registers force perpendicular to the horizontal. So the scale only measured force perpendicular to the lever arm.
I'm not saying my results aren't wrong, i trust Lino's pressure gauge more than my bathroom scale. I'd just like to know how i got 12 bar.