I hadn't realized that most spreadsheet programs can do multiple regression until recently. Using that capability, McMaster-Carr's reported LRT pressures and the knowledge that the reported pressure seems to be for the initial pressure spike and that the operating pressure is about 2/3 of the reported, I've made a relatively simple equation that describes LRT pressure as a function of tube dimensions.

P(t,d_i) = 31.4 * sqrt(t) + 18.0 * (t / d_i)

where

P is pressure in psi,

d_i is the internal diameter of the tube (unfilled) in inches, and

t is the tube wall thickness.in inches

Note that the relationship is not quite that simple as the tube's pressure will depend on how full the tube is, the number of cycles the tube has gone through, the age of the tube, temperature, among other things I'm probably not aware of. It's a good start however and will suffice until I figure out a relationship based upon a theory of rubber and do tests of my own.

If I find a better relationship or one that makes more intuitive sense I'll post it.

Edit:

The equation below used to be posted but I replaced it with a simpler one that I think is more representative of what's occurring.

P(t,d_i) = 5.85 - 3.38*d_i + 57.98*t + 15.83*(t/d_i)

## LRT pressure as a function of tube dimensions

- SSCBen
**Posts:**6449**Joined:**Sat Mar 22, 2003 1:00 pm

### LRT pressure as a function of tube dimensions

Last edited by SSCBen on Fri Apr 17, 2009 1:57 am, edited 1 time in total.

- Silence
**Posts:**3825**Joined:**Sun Apr 09, 2006 9:01 pm

### Re: LRT pressure as a function of tube dimensions

What type of variance and/or standard deviation does that regression produce? Can you plot your regression curve on a scatter plot? I trust that the regression is accurate, so this looks like a pretty useful formula.

You might want to mention that the lengths are in inches.

You might want to mention that the lengths are in inches.

- SSCBen
**Posts:**6449**Joined:**Sat Mar 22, 2003 1:00 pm

### Re: LRT pressure as a function of tube dimensions

The "standard error" is the average standard deviation for the entire curve. The variable for the standard error is s in the notes I have.

s = 3.1 psi

R^2 = 0.9847 (for the more regression with full precision, not the simplified formula I posted)

n = 37

Attached is the spreadsheet itself for those who want to take a closer look at the data.

Remember that this equation is only as accurate as the data and the assumptions used. McMaster-Carr has precision of only 5 psi on some tubes and some of the values seem inconsistent. The 2/3 multiplier to convert from reported to adjusted pressure is mostly arbitrary and made sense when I did some tests with one tube. It may not be correct. More research and a correct equation based on the theory of rubber elasticity (which is based on entropy as far as I know) is necessary, but this is very adequate given the lack of an alternative.

Also attached is a screenshot of gnuplot that should help show how close the regression is to the data. I'd suggest using a 3D plotting program if you'd like to examine things like this because a simple screenshot does not give you the full picture.

s = 3.1 psi

R^2 = 0.9847 (for the more regression with full precision, not the simplified formula I posted)

n = 37

Attached is the spreadsheet itself for those who want to take a closer look at the data.

Remember that this equation is only as accurate as the data and the assumptions used. McMaster-Carr has precision of only 5 psi on some tubes and some of the values seem inconsistent. The 2/3 multiplier to convert from reported to adjusted pressure is mostly arbitrary and made sense when I did some tests with one tube. It may not be correct. More research and a correct equation based on the theory of rubber elasticity (which is based on entropy as far as I know) is necessary, but this is very adequate given the lack of an alternative.

Also attached is a screenshot of gnuplot that should help show how close the regression is to the data. I'd suggest using a 3D plotting program if you'd like to examine things like this because a simple screenshot does not give you the full picture.

Last edited by SSCBen on Sat Apr 18, 2009 4:14 pm, edited 1 time in total.

- cantab
**Posts:**1492**Joined:**Fri Oct 19, 2007 1:35 pm

### Re: LRT pressure as a function of tube dimensions

IMHO that plot is meaningless. The points could be anywhere, since it's a 2d projection of a 3d space.

You need something like rods, going up from P=0, to make the data point locations unambiguous.

You need something like rods, going up from P=0, to make the data point locations unambiguous.

I work on Windows. My toolbox is Linux.

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Arsenal:

Super Soaker: XP215, 2xXP220, Liquidator, Aquashock Secret Strike M(odded), Arctic Blast M, CPS1200, CPS2100, SC Power Pak, 3l aquapack, 1.5l aquapack

Water Warriors: Jet, Sting Ray M, Shark, Argon M, Tiger Shark, PulseMaster

Others: Waterbolt, The Blaster, Storm 500, Shield Blaster 2000, generic PR gun, generic backpack piston pumper (broken), 3l garden sprayer M, 10l water carrier:

- SSCBen
**Posts:**6449**Joined:**Sat Mar 22, 2003 1:00 pm

### Re: LRT pressure as a function of tube dimensions

I agree that the plot is mostly worthless. I think gnuplot can do precisely what you've described but I don't know the specifics. The data's available so if someone wants a more precise comparison, one can be made.

Also, this relationship suggests that a 50 psi tube with an ID of about 0.75 inches should have a wall thickness of about 1 inch. While I'm not sure of the accuracy of this formula beyond the limits of the input data, this demonstrates the pressure weakness of LRT. I crunched the numbers for a bunch of tubes available from a manufacturer and no practical tube had a pressure of over 40 psi.

Also, this relationship suggests that a 50 psi tube with an ID of about 0.75 inches should have a wall thickness of about 1 inch. While I'm not sure of the accuracy of this formula beyond the limits of the input data, this demonstrates the pressure weakness of LRT. I crunched the numbers for a bunch of tubes available from a manufacturer and no practical tube had a pressure of over 40 psi.