Converting Pascal to PSI Trying to do a basic unit version and getting stuck. Also, I tried formatting this with MathJax or whatever it's called but it looked terrible.  Couldn't get it to look right so if anyone can assist with that that would be appreciated.
A Pascal is defined as $\frac{\mathrm{kg}}{\mathrm{m} \, \mathrm{s}^2}$. When doing the conversion, I get close to the final answer, but not quite there.
1 Pa = $\frac{\mathrm{kg}}{\mathrm{m} \, \mathrm{s}^2} \cdot \left(\frac{1 \, \mathrm{lb}}{0.4536 \,\mathrm{kg}}\right) \cdot \left[\frac{1 \,\mathrm{m}}{39.37 \,\mathrm{in}}\right]^2$
I then end up with: $0.0014 \, \frac{\mathrm{lb} \, \mathrm{m}}{\mathrm{s}^2 \, \mathrm{in}^2}$
I know this is close to the final answer, as the actual conversion from Pascal to PSI is 0.00014, so I am off by a factor of 10.  What am I missing??
 A: Converting kilograms to pounds, even though useful in a grocery store, is not a valid operation from point of view of dimensions. One is mass, the other is force. Newtons to pounds will be ok and give you the right answer.
A: Convert Pa in $\rm{kg\over {m\sec^2}}$ to $\rm{lbm \over {ft\sec^2}}$ then use $\rm{1\ lbf = 32.174\ {{lbm\ ft} \over sec^2}}$ to obtain pressure in $\rm{lbf \over ft^2}$, then convert to $\rm{lbf \over in^2}$.
Note: in English units $\rm{1\ lbf = 32.174 \ {{lbm\ ft} \over sec^2}}$.
A: Dimensions of Pascal
Pascal is the SI unit for pressure, which is force per unit area.  In SI units, this corresponds to $N/m^2$.  To get to SI base units, note that Newton's 2nd law can be used.  Thus, the following are the base SI units for a Pascal:
$Pa = \frac{N}{m^2} = \frac{ma}{L^2} = \frac{kg-m/s^2}{m^2} = \frac{kg}{m-s^2}$
Dimensions of psi
psi is also a pressure term, so it represents a force per unit area.  The units of psi are $\frac{ib_f}{in^2}$.  Note the subscript "f" on "lb".  This represents pounds force, which is distinctly different than pounds mass.  Again, use Newton's 2nd law to arrive at the units of pounds force:
$lb_f = lb_m * 32.2 \frac{ft}{s^2}$
Recommendations for unit conversions
There are several practices that greatly reduce the chance of errors when doing unit conversions.  Those practices are:

*

*Draw square brackets and draw a horizontal division bar between those brackets.  It is important to draw a HORIZONTAL division bar rather than a slash, because the horizontal bar makes it much easier to determine whether or not units cancel later on in the procedure

*Write the units that you want to convert inside the brackets just drawn

*Assuming that there are units in the numerator and denominator, choose one to convert

*Draw the next pair of square brackets, with a horizontal division bar

*Write UNITS first in the numerator and the denominator, ensuring that units cancel properly between the numerators and denominators of your brackets

*Now, write the numbers that match the units in the bracket you just drew.  For example, if your 2nd bracket has units of $\frac{m}{ft}$, you would write 1 in the numerator and 3.28 in the denominator.  You will know that the numbers are correct when you check the ratio in the brackets.  That ratio should equate to 1, and the ratio of 1 m divided by 3.28 ft does indeed equal 1 because 1 m equals 3.28 ft

*Continue with brackets until all units in the numerator and the denominator are converted to the units that you want

*Make sure that you lightly cross out all of the units that cancel, and inspect the result to ensure dimensional consistency.  For example, if you have ft in the numerator of one factor and one other occurrence of ft in the denominator of any factor, $\frac{ft}{ft}$ equates to 1, and it cancels out of the problem

Converting Pa to psi
Using the recommendations above, note that the units of Pa are:
$[\frac{kg}{m-s^2}]$, and that is the first factor to write.  Starting with the numerator, it is seen that the units are kg, which is a mass, and this needs to be converted to $lb_f$ in order to arrive at psi.  Thus, the complete set of factors required to convert 1 Pa to psi is as follows:
$[\frac{1 kg}{m-s^2}][\frac{2.2 lb_m}{1 kg}][\frac{1 lb_f-s^2}{32.2 ft-lb_m}][\frac{3.28 ft}{1 m}][\frac{1 m}{39.37 in}]^2 = 0.00014458 psi$
A: It would be easier if you treat pascals as "Newtons per square meter". PSI is "Pounds per square inch", right? Well, both inches and meters are units of length, and both newtons and pounds are units of force. So, try to convert the newtons to pounds and meters to inches using a simple conversion factor. Dont lay out all the units kg m * s^2. It's easier to do it the other way. Hope this helped! :) Im new here by the way, just signed up 5 min ago. :D
A: The Pascal is, as others have answered, defined as $1$ $N$ acting on a surface of $1$ $m^2$.
$$N = ma$$
$$a = g$$
$$g \approx 10 \frac{m}{s^2} (1)$$
(1) earth's gravitational constant
And there is your missing zero. 1 Pascal is $0.1\frac{kg}{ms^2}$ if you wish to translate to "kilograms of force on the surface of the earth".
