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Not being a math major or engineer, (I'm a warranty manager), is there a formula to calculate the PSI exerted on the cheese using a lever press.  How would one calculate the amount of weight required on the end of the lever to exert the required psi on the cheese plunger?

Thanks

You're talking about a couple of different concepts. It helps to separate them, divide and conquer, eh.

First the PSI. It's Pounds per Square Inch. It's a pressure unit.

Your press will deliver a specific number of pounds on the follower, and the follower will have an area. Those two combined will give you a pounds per area, and if your area number is calculated in square inches, then you'll have your PSI.

A circular area is calculated using the formula A= pi * (radius squared)

pi is 3.14159

the radius of a circular area is measure from the middle to the outside, or you can use the maximum width of the circle (the diameter) and divide by 2.

So for a 4inch mold,

A = 3.14159 * (radius of 2inches) squared
A = 3.14159 * 2 squared
A = 3.14159 * 4
A = 12.57 square inches

So, you can calculate the area of the top of your mold, now onto the weight

I haven't seen the design of your press, but I assume that it's using a lever to magnify the downward force with one end being fixed.

We'll tackle that one next.

There is probably a main horizontal lever that your design has.

On (or towards) one end is the weight, and the other end is fixed. Somewhere between those two is the place where force is exerted on the follower.

There are two distances you'll be concerned with.

1) The distance between the weight and the fixed end, and
2) The distance between the follower connection and the fixed end.

The force being exerted at the follower in pounds will be the weight in pounds (times) distance #1 (divided by) distance #2

So if you hang a 50lb weight on the end, and distance 1 is 2meters and distance 2 is 1 meter, then the force on the follower is 50 * 2 / 1 = 100lbs.

That 100lbs on your 12.57sq inch follower would be 7.9PSI on the cheese.

Good job, Groves. 

Here is a normal lever type diagram, in the case of a cheese press the "Output" is at the base of the fulcrum.
Grove's excellent description has distance 1 to the left of the fulcrum and the weight or work end to the right.

I believe that most leverage presses will have the fulcrum at the end, and both the input and output will be away from the end, but the numbers would work the same.

More like this, excusing the fact that this isn't a press at all, but illustrating a common configuration.

Whew, thanks for the replies.  Now if I read this correctly:

My hoop is 7" in diameter so the radius is 3.5 correct?

A = 3.14159 * (radius of 3.5inches) squared
A = 3.14159 *3.5 squared
A = 3.14159 *12.25
A = 38.46 square inches

So if my weight is 28" to the fixed point, and my follower is 10" to the fixed point, then 28/10 = 2.8 * weight hanging.  Such as:  30 lbs x 2.8 = 84 lbs

So 84/38.46 = 2.18 lbs of pressure per square inch?

So to have 50 lbs of pressure on the cheese in the 7" hoop I would reduce the hanging weight to about 18lbs correct.

Thank you Groves and Coffee Joe for your assistance in this.  Please correct me if I have miscalculated.  I really appreciate this help.  I do not have a press yet, but it is being built this week.

kim

QuoteSo to have 50 lbs of pressure on the cheese in the 7" hoop I would reduce the hanging weight to about 18lbs correct.

This is true but is this what you need? The psi would only be between 1 and 2 lbs/sq in. I am not sure what kind of cheese you are making but for some cheeses (cheddar, jack, colby) you need more pressure than this. You may already understand this, so please ignore if you do. Many times the recipes will just give a required weight and I wanted to be sure you really understand that the 50 lbs on the cheese will produce varying pressures dependant on the area/ radius of the hoop. Your math is great! You even showed your work. 

Here's a handy little table that shows the weights and corresponding psi for various hoop sizes. I have this laminated and keep it by my press for reference.

Find your hoop size at the top. Go down the column until you find the psi that you want to press your cheese at. Now follow the row to the left to find the weight that you need to produce that psi.

Example: You are doing a Gouda in a 7" hoop so find that at the top. A good pressing for Gouda is 2 psi, so go down the column to 2 (or close to it). To the left you will see that it takes 75-80 pounds to hit that psi target.

To put psi into a practical perspective...  When using a 4" hoop, it only takes 25 pounds to achieve the same psi. Recipe books apparently assume that home cheese makers are using 4" hoops. They do NOT tell you how to adjust for larger cheese sizes.

Sailor, I do not see the chart.  I appreciate your comments too.  Thank you.  I would love to have that chart.  I just made a 3 gallon cheddar, and after pressing it is only about 2 1/4" tall.  Very thin cheese.  Once the rind develops, there will not be much cheese.

FarmenrJd - thank you as well.

I think I need to go to a deeper and small diameter hoop.  What do most of you use for a hoop standard size?

There was quite a discussion of this previously:

https://cheeseforum.org/forum/index.php/topic,4026.msg30656.html#msg30656

I have included my latest calc sheet for my press as an additional data point for you. 

-Boofer-

Thanks Boofer, I will definitely read that thread.  Thanks for the chart too.  It's on my desktop.

I know I attached the chart last night - Don't know what happened.

Anyway, here it is again in both Acrobat and Excel formats.

Excellent chart Sailor,

Now, I'm always in doubt as to the proper PSI for different cheeses, Cheddar, Parm, colby etc

I was warned that the PSI I was using for cheddar was low so I increased my arm (MA) to 5.22 with 6Kg and 22 Kg on 8" mold

so am now running 1.43 PSI for 20 Min, then Flip over in mold and go to 5.23 PSI for 24 Hours, change cloth and turn over at same weight for another 24 hours.

I have a single pulley on my press( not using) as well so that would double the PSI

I think I did this math correctly