How many psi's are in one bar?

6894.7573 bar = 100000.0 psi according to google

6894.7573 bar = 100000.0001 psi according to wolfram alpha

which is it? How many psis are in one bar? Is there an exact value?

$$\frac{100,000\,{\rm psi}}{ (4.4482216152605\,{\rm N} / (0.0254 \,{\rm m})^2 )} = 14.503773773020921515424102795119\,{\rm \frac{psi}{bar}}$$

.. which seems to conflict with some answers and some wiki pages. So I am looking not just for the value, but for some significant authoritative source, or derivation from said sources.

• So that 1 in the 4th decimal place is bothering you? – Kyle Kanos Nov 7 '14 at 19:12

Sometimes a picture tells a thousand words...

It all depends what question you ask Wolfram Alpha: Floating point arithmetic leads to rounding errors. Non SI units are rarely defined precisely (an exception is the inch which is exactly 25.4 mm - and thus other derived units of length).

But getting back to the "what is the value" - we should go with actual units that have defined quantity (rather than the ones that are "precise to a certain number of digits"):

The pound (according to the National Bureau of Standards is defined as:

1 pound (avoirdupois)= 0.453 592 37 kilogram defined - exact

The inch is defined as

25.4 mm exact

The bar is defined as:

1 $N / m^2$ exact

The relationship between Newton and kilogram:

g = 9.80665 $m / s^2$ defined - "exact"

Now, all you need is a calculator with infinite precision, and you can compute your number to your heart's content...

I get for 100000 psi:

100000 * 0.45359237*9.80665 / (10 * 2.54 * 2.54) =
6894.757293168361336722673445346890693781387562775125

and for 1 bar:

2.54 * 2.54 * 10 / (0.45359237*9.80665) =
14.50377377302092151542410279511939767008919516410

comparing to your value:

14.503773773020921515424102795119

As you can see, the value you got was "accurate" to all the figures you had. But I have more...

There is an often quoted paper about "What every computer scientist needs to know about floating point arithmetic". Worth reading: http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html

I say this often: you should be very, very wary of any situation in which you need to know a quantity to better than 1%, 0.1%, 0.01%... Quite likely you are tackling the problem in the wrong way and there's an elegant method waiting to be found. Except as an intellectual exercise, finding out the Nth digit of a physical quantity that isn't even an SI unit is silly. I wrote this answer just to point out that if you go back to sources, you can get the answer.

Now that's a bit silly, but since each of these values was defined, this is in principle accurate to all these (insignificant) digits. I use the word "accurate" very, very loosely.

• thank you, that is quite an interesting answer worth reading. I am glad there was a way to go back to sources on this. And +1 for "accurate" to all the figures you had. But I have more... – Dennis Nov 10 '14 at 14:58
• and indeed I was surprised that for engineering applications (field I am writing code for), they accept precision for up to 3 decimal places as good enough. – Dennis Nov 10 '14 at 15:02
• @Dennis - since engineering typically uses "safety factors" that may be 3x, 4x, 5x of the computed values, you really don't need more than 3 digits for most calculations. Of course during finite element analysis etc, you are taking differences between numbers and you need to be very careful about sig figs. – Floris Nov 10 '14 at 18:04

1 PSI is 0.0689475728 bar which means that 1 bar is 14.50377397 PSI. Thus, $$\frac{6894.7573\,{\rm bar}}{1}\times\frac{14.50377397 \,\rm PSI}{1\,\rm bar}=100000.0002900755\,\rm PSI$$ which is slightly off from both sources.

NIST says that

• 1 Bar = $10^5$ Pascal
• 1 PSI = $6.894757\times10^3$ Pascal

Thus, $$6.894757\times10^{-2}\,{\rm Bar}=6.894757\times10^3\,{\rm Pascal}= 1\,{\rm PSI}$$ which is slightly fewer decimal places than the 0.068975728 I cite above. This would give, $$6894.7673\,{\rm Bar}=100000.0044\,{\rm PSI}$$