I started a new thread... Dan can merge it with "Rescuing an '83..." if he sees fit...

Looking at finding a rep. spring for a '83 Microcasa...

I've heard that the Elektra starts at 8 bars and finishes around 6. Can anyone verify?

I'm looking at springs here and that's on the shady side of possible, but only if you start with a spring that's about 24" long (OK I'm exaggerating)...

Let's look at the microcasa.
travel is just over 1", call it 1.0
spring free length is 2.7" removed from machine (on Bob's tired spring)
spring in machine (lever up, minimally compressed) 2.3"
and compressed is 1.3"
bore diameter is 1.563" area is 1.92 sq in
(all numbers are near, not exact, good enough for this illustration)

Now lets look at spring rates:

To finish at 6 bar, or 90 psi the lever up force needs to be 172#
uncompressed that force must be zero, so you get 172# in .4", or a spring rate of 431#/in.
that gives a lever down force of 603# or a pressure of 314 psi or 20 bar!

Let's backtrack the other way.
8 bar start is 120 psi or 230#
6 bar finish is 90 psi or 172#
1" travel between gives 57# per inch.

To get to a 6 bar finish the free (out of machine) spring length must be 5.3" to give the 3" pre compression necessary for 172#

It's not very easy for a spring to go from 5.3" to 1.3" without bottoming out...

Anyone seen an uninstalled (new) microcasa spring?

Also the '83 parts look identical to the Microcasa vs Pavoni pics (I'll post some later)

ciao

lino

Lino,

I took am looking for a replacement spring... and I'm frustrated with the wimpy spring on my 1993 Elektra. Jimmy wants 8.5+ bars!

Any luck?

-jimmy

I'm posting in this topic just to keep this information together, hope somebody can answer this question!

What is the avg. life expectancy of a spring? Is it something that should be replaced just like gaskets?

The reason I'm asking is 'cause the spring action of my 26 year old (2nd hand) Zerowatt is very weak. It takes very little effort to depress the lever and it doesn't develop anything near the pressure needed to get crema.

I've been getting great drinks by using a coarser grind and light tamp, but I'd like to replace the spring. I'm a bit worried that replacing it with a stronger spring might put too much stress on other components.

Anybody have experience with this?

Henry

When computing the strength of the spring required to produce a target pressure (Force/Surface Area) would one use the surface area of the puck? (i.e. PI * Basket Radius squared? )

Regards
Timo

The pressure is the same anywhere under the spring to the top of the puck (just like a hydraulic pump which changes crossection area to get mechanical advantage) -- so one only needs to worry about the cylinder's diameter for pressure calculations, not the puck or basket.

The spring is equivalent to a weight on the cylinder (and I don't want to see any pictures of groups with barbells instread of levers!). The brewing pressure is the weight divided by the cylinder area.

another_jim wrote:The spring is equivalent to a weight on the cylinder (and I don't want to see any pictures of groups with barbells instread of levers!).

(apologies for the noise...)

Is the force required to compress a spring equal to the strength with which it presses back?

Henry

another_jim wrote:The pressure is the same anywhere under the spring to the top of the puck (just like a hydraulic pump which changes crossection area to get mechanical advantage) -- so one only needs to worry about the cylinder's diameter for pressure calculations, not the puck or basket.

The spring is equivalent to a weight on the cylinder (and I don't want to see any pictures of groups with barbells instread of levers!). The brewing pressure is the weight divided by the cylinder area.

But not if the piston has a different diameter than the basket, right? Are you assuming they're the same diameter?

When the diameter of the basket is smaller than the diameter of the piston, I think you should use the diameter of the basket when calculating the pressure. Or if the diameter of the basket is larger than the diameter of the spring, you should use the basket too.

Pressure = Force / Surface Area.

An analogy.

If a large stereo speaker weighing 60 pounds (let's say it's 1ft wide by 1ft deep) is resting flat on the floor, the weight of the speaker is distributed over 1 square foot. The pressure on that 1 square foot of floor is 60 pounds. 60 pounds divided by 144 square inches. The pressure on any one of those square inches in that scenario is 60/144.

But now let's take the same speaker and add four tiny feet, inverted cones designed to minimize vibration. The cones are .25" square (one quarter-inch squared) where they meet the floor. So now we have 60 pounds distributed over 1 square inch (there are four of these feet) so that now one square inch is bearing the full 60 pounds of weight. The pressure on the square inches of floor where the feet touch is 60 pounds, not 60/144.

Same weight, different surface area. Different pressure.


Regards
Timo

A hydraulic pump (from grade school physics):

Take a 1 inch cross-section tube, press on it with 1 pound -- have it make a u-turn and flare out to 100 square inches. The water will be at 1 psi throughout the system, and the 100 sq inch side will lift 100 pounds. Obviously, you have to press down 100 inches to lift the other side 1 inch.

The pressure is equal throughout the sealed part of the system, cylinder bottom to puck top. So that pressure will be the force on the cylinder divided by its surface area, regardless of what happens lower down. A narrow cylinder means more piston travel for the same amount of coffee, but less force -- the total amount of energy required is the same.

another_jim wrote:A hydraulic pump (from grade school physics):

Take a 1 inch cross-section tube, press on it with 1 pound -- have it make a u-turn and flare out to 100 square inches. The water will be at 1 psi throughout the system, and the 100 sq inch side will lift 100 pounds. Obviously, you have to press down 100 inches to lift the other side 1 inch.

The pressure is equal throughout the sealed part of the system, cylinder bottom to puck top. So that pressure will be the force on the cylinder divided by its surface area, regardless of what happens lower down. A narrow cylinder means more piston travel for the same amount of coffee, but less force -- the total amount of energy required is the same.

(my emphasis)

I must have played hookey that day, Jim. Please bear with me.

When you write "its surface area" I take it you mean the cylinder's surface area? How would you compute "the surface area of the cylinder"? Would you measure the surface area of the inner wall plus the surface area of the top and bottom circular planes? Or you would you measure the surface area of a cross-section of the cylinder?

Target Pressure stated in PSI = Force Applied in Pounds / Surface Area in Square Inches

Regards
Timo

Just in case you aren't pulling my leg: