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Community Tip - Help us improve the PTC Community by taking this short Community Survey! X  1-Newbie

I am using MathCAD Prime 2.0 for design calculations for a building I am engineering. I am fairly new to MathCAD, only learning a little about it in college. However, I am running into a units issue when calculating pounds per linear foot (lb/ft) to just pounds (a point load). I am taking 480 PLF * 4 ft, the feet should cancel out and you are left with just pounds. MathCAD is spitting out a lb*f, like the units for a moment. When I go to change the units to just pounds then it changes my units to (ft/s^2)*lb? Please explain to me what is going wrong here. See below for a screen shot: Please note that the answer is correct, 1920 pounds is the right answer. Just the wrong units.

Jon Recknagel | Structural Designer

1 ACCEPTED SOLUTION

Accepted Solutions  1-Newbie
(To:ptc-4992678)

I just answered my own question. A "lbf"= a pound force. It seems like they could have just left it at "lb" we are all aware that pounds is a force. Just my opinon I guess. Thanks anyway.

20 REPLIES 20  1-Newbie
(To:ptc-4992678)

I just answered my own question. A "lbf"= a pound force. It seems like they could have just left it at "lb" we are all aware that pounds is a force. Just my opinon I guess. Thanks anyway.  19-Tanzanite
(To:ptc-4992678)
 It seems like they could have just left it at "lb" we are all aware that pounds is a force.

The pound is a unit of mass, not force.  1-Newbie
(To:RichardJ)

I would disagree....Pounds (lb) is not a mass. Pounds is a force. F=mg, due to gravity.  19-Tanzanite
(To:ptc-4992678)

What unit are you using for m?  1-Newbie
(To:RichardJ)

m=F/G, most engineering firms (in the US anyway) do not work in terms of metric units of mass. Everything is in pounds, a force. When you step on the scale at home you do not read your weight in terms of mass, you read your weight it terms of pounds, due to gravity, which is a force.  19-Tanzanite
(To:ptc-4992678)

It's not a question of metric vs non-metric. I asked because the only unit systems in which pounds are a force are the gravitational systems, in which the base units are force, length, and time, rather than mass length and time. If the base unit of force is the pound, then the mass unit is the slug, and is defined in terms of the pound. The slug is defined as a mass that accelerates by 1 ft/s2 when a force of one pound is applied.

All unit systems in Mathcad are based on mass, length and time though. It does not have unit systems based on either force, length and time as the base units, or weight, length, time as the base units. In Mathcad a pound is always a mass. A pound force is distinguished from that by using the nomenclature lbf.

What a bathroom scale says is irrelevant. Most people don't even know the distinction between mass and weight. You can get scales that read kg too, and there is no unit system in which kg is a unit of anything other than mass.  1-Newbie
(To:RichardJ)

"lbf" is a unit of force.

In the English unit the correct mass is a slug but no body uses the unit.  23-Emerald I
(To:MS_8437498)

You are absolutely correct!

lbf is pound force.  lb (or lbm) is mass, so is slug.   5-Regular Member
(To:Fred_Kohlhepp)
Yep. Mathcad allows definition of lbm and lbf, for pounds of mass and pounds of force. I use these everyday when doing refrigeration work. Values for enthalpy (in the I-P system) are in BTU/lbm, but I also have to convert to psi for pipe pressure loss calculations. The gravitational constant is the method I use of converting from one to the other. The gravitational constant can be expressed as g_c = 32.174 ft*lbm/lbf*s..Use this in a calculation of the Generalized Energy Equations (derived from the Bernoulli equation) and it will allow use of pressure in psi and pounds of mass in the same equation. Also, the gravitational constant is equal to zero, so anything can be multiplied by it.  23-Emerald I
(To:SL_9674672)

I thought there was something wrong here!    5-Regular Member
(To:Fred_Kohlhepp)
Reference this paper; Piping-System Solutions Using Mathcad, B.K, Hodge and Robert P. Taylor, May 1, 2002. Without the use of g_c, this methodology won't work in Mathcad and would require iterative hand solutions. Close enough is better than by hand. Also, you are likely seeing an error from defining g_c to only 3 decimal places. Extend your allowable decimal places to 12 for the definition of g, then use that to define g_c and it will be nearer to zero. Later today I should be able to post an actual example, gotta do some software maintenance first.  23-Emerald I
(To:SL_9674672)

Okay, attached (in Prime 4 Express) is a start through the paper Piping-System Solutions Using Mathcad, B.K, Hodge and Robert P. Taylor, May 1, 2002.

By doing a unit balance of equation (1), we can correct the equation and demonstrate that the conversion factor is NOT required.  12-Amethyst
(To:Fred_Kohlhepp)

Dear Fred, could you attach also a pdf? Thank you so much. Filip  23-Emerald I
(To:FDS)

Original paper an mathcad file pdf attached  5-Regular Member
(To:Fred_Kohlhepp)
I really have to talk to our IT guys and find out why I can't post a file. On the second page, W.s = g/g.c * z.a, The gravity term is not needed. That is why the result has an additional acceleration. The g.c is needed so the Ws comes out in feet. The equation is in terms of feet, especially when related to a hydronic system. Perhaps it is not required, but conversion to a pressure system vs. a pressure related to column height, is not really required. But when I analyze a refrigeration system, I am looking at flow rates in pounds of mass. Converting these flow rates to equivalent pressure losses is absolutely required. Also, all unit conversions are essentially multiplication by 1. They are never necessary but always helpful. Never required but performed constantly. When using IP units, conversion factors are absolutely necessary, unless we want to brush to the dust off the old slugs. The units are handled in one way or another. I don't think it is any easier to express or apply than using g.c.  23-Emerald I
(To:SL_9674672)

You clearly have a system that works for you; enjoy it and good luck.

I spent (roughly) twenty years doing this type of calculation (and others) in Mathcad and wrestling with units.  Much of that effort was spent taking expressions like equation (1) and making sure the units balanced properly; That's what I was attempting here.  "all unit conversions are essentially multiplication by 1. They are never necessary but always helpful."  One of Mathcad's features that has saved me numerous times is the automatic unit conversions.  If I expect an calculation to result in a moment (ft lbf) and the answer Mathcad labels J, I can simply replace Mathcad's units with mine: and have the units I want.  If I screw up the input, Mathcad balances the units , and I can see there's a problem.

With that in mind I propose a new pressure unit   5-Regular Member
(To:ptc-4992678)

The "lbf", pounds force is there to differentiate between pounds mass, which is just "lb". Something which matters greatly when doing dynamics, rather than statics.  24-Ruby V
(To:ptc-4992678)

You may want to check how the different units for Force per Length are defined in Mathcad   23-Emerald I
(To:Werner_E)

Careful!

You've run afoul of Mathcad's rule to use only the units required.  The definitions are valid if you proscribe the units   23-Emerald I
(To:Werner_E) 