Question about water pressure in oceans:

Originally posted by sonhouse
The latter, by what, a thousand times. I guess you could rate pressure in its ability to punch holes in the object being touched.

Actually, its greater by 1 million times.

Originally posted by joe shmo
Actually, its greater by 1 million times.

That was my first choice but I edited it out to 1K. So it goes up by mm^2.

Originally posted by sonhouse
That was my first choice but I edited it out to 1K. So it goes up by mm^2.

Correct, the pressure scales with the area. Start with (1)

1[m²]= 1[m²] (1)

Then manipulate one side of the expression by multiplying it by factors of one

eg. (1000[mm]/1[m]) = 1

So that what you have is

1[m²] = 1[m²]*1*1

1[m²] = 1[m²]*(1000*[mm]/[m])(1000[mm]/[m])

Multiply the factors

1[m²] = 1[m²]*(1000*[mm]/[m])²

1[m²] = 1[m²]*1000²*[mm²]/[m²]

As you can see the on the right side of this expression m² in the numerator will divide out with m² in the denominator, leaving you with the appropriate unit transformation.

1[m²] = 10^6 [mm²]

Originally posted by joe shmo
Correct, the pressure scales with the area. Start with (1)

1[m²]= 1[m²] (1)

Then manipulate one side of the expression by multiplying it by factors of one

eg. (1000[mm]/1[m]) = 1

So that what you have is

1[m²] = 1[m²]*1*1

1[m²] = 1[m²]*(1000*[mm]/[m])(1000[mm]/[m])

Multiply the factors

1[m²] = 1[m²]*(1000*[mm]/[m])²

1[m²] = 1 ...[text shortened]... in the denominator, leaving you with the appropriate unit transformation.

1[m²] = 10^6 [mm²]

So to my question about what the pressure of a vinyl stylus:

https://sites.google.com/site/zevaudio/turt/stylus-shape-information

It lists contact from 20 to 62 microns. Can you tell by the bottom image if that is 20 to 60 SQUARE microns or not? I would think it was. The piece says the larger contact gives the longest lasting needle and the smallest pressure.

I am assuming 20 to 60 micron ^2, with one gram of weight, what is the actual pressure on the record?

I have high tech recording equipment, both analog and digital but still have 2000 vinyls, mainly classical and folk music so I am interested in keeping my collection in the best shape possible.

Didn't know about all those different stylus shapes, except conical vs elliptical.

Maybe I will get one of the S.A.S. ones if I can find them maybe on Ebay.

Wow. I found them, Shibata stylus with cart., 200 to 600 bucks each!

Originally posted by sonhouse
So to my question about what the pressure of a vinyl stylus:

https://sites.google.com/site/zevaudio/turt/stylus-shape-information

It lists contact from 20 to 62 microns. Can you tell by the bottom image if that is 20 to 60 SQUARE microns or not? I would think it was. The piece says the larger contact gives the longest lasting needle and the smallest ...[text shortened]... find them maybe on Ebay.

Wow. I found them, Shibata stylus with cart., 200 to 600 bucks each!

Yeah, It states contact surface, which means Area. So I'll break this up, and you can do the math.

We first need to find the weight of the stylus in Newtons [N].

Mass = 1 gram
gravitational acceleration = 9.81 m/s²

Weight = Mass*gravitational acceleration

1 N = 1 kg*m/s²

So what is the weight "W" of a mass of 1 gram on earth in Newtons [N]

Next we need to convert square microns to square meters, Ill get you started

20[mic²] = 20[mic²]

Multiply the expression above by the appropriate factor below, the correct number of times such that the [mic²] will divide out leaving you with [m²].

Factors:
(1,000,000[mic]/1[m]) = 1 or (1[m]/1,000,000[mic]) = 1

Then all that is left is to divide the weight ( that I hope you calculated ) with the Area in [m²] and out comes the pressure in Pascals[Pa]

And typical atmospheric pressure is 101,300 [Pa] so how many times greater is this pressure relative to atmospheric pressure?

Originally posted by joe shmo
Yeah, It states contact surface, which means Area. So I'll break this up, and you can do the math.

We first need to find the weight of the stylus in Newtons [N].

Mass = 1 gram
gravitational acceleration = 9.81 m/s²

Weight = Mass*gravitational acceleration

1 N = 1 [b]kg*m/s²

So what is the weight "W" of a mass of 1 gram on earth in Newton ...[text shortened]... re is 101,300 [Pa] so how many times greater is this pressure relative to atmospheric pressure?[/b]

If I did this right, I think it comes out to 2.5E6 pascals? 24 atmospheres or 362 pounds per square inch? I am basing that on 20 square microns means you can put 2500 such units in one mm^2 which would mean 2.5E9 such units in a square meter then divide by 1000, gives you 2.5E^6 pascals. Right?

So if that is right then the one called S.A.S at about 60 microns would come out to 1/9th the pressure which is why it would last so much longer both for record and stylus lifetime.

Originally posted by sonhouse
If I did this right, I think it comes out to 2.5E6 pascals? 24 atmospheres or 362 pounds per square inch? I am basing that on 20 square microns means you can put 2500 such units in one mm^2 which would mean 2.5E9 such units in a square meter then divide by 1000, gives you 2.5E^6 pascals. Right?

So if that is right then the one called S.A.S at about 60 mi ...[text shortened]... /9th the pressure which is why it would last so much longer both for record and stylus lifetime.

I get 4.905E8 pascals. So something is off on your calculation.

Originally posted by joe shmo
I get 4.905E8 pascals. So something is off on your calculation.

A mere 2 1/2 orders of magnitude. Almost right on🙂 Have to go back to the drawing board.