# Force of Lift Unit Conversion?

1. Oct 30, 2009

### aaronmorg

I am stumped on how to convert my units for the equation for Force of Lift.

FL=CLPA$$\frac{v^{2}}{2}$$
My equation comes out to:

FL=(.76764)(.076$$\frac{lbm}{ft}$$)$$(\frac{(65.899\frac{ft}{s})^{2}}{2})$$

I am supposed to get lbf from all this. I think.

2. Oct 31, 2009

### FredGarvin

There are two things you can do in a situation like this:

1) First convert all of the units to SI units and rework the equation. There is no confusion on units in the SI system. Then when you are done, convert back over to US units.

2) Look at the density term. The standard units for density in our system is slugs, not pounds mass. That will give you
$$\frac{sl*ft}{s^2}$$ which is Lbf

or just multiply your answer by 32.2 and trust that the units turn out right (which they do)

This may help:
http://en.wikipedia.org/wiki/Pound-force

3. Oct 31, 2009

### Staff: Mentor

....and thats slugs per cubic foot!

Last edited: Oct 31, 2009
4. Oct 31, 2009

### FredGarvin

Oh yeah. Look at that.

What Russ said...

5. Nov 1, 2009

### aaronmorg

Thanks :D

I'll go try converting to metric units and I'll let you know how everything turns out!

I appreciate the help.

6. Nov 1, 2009

### Staff: Mentor

Since you're going with this method, just a little comment: Though the units of SI are easier to work with, converting to SI does not eliminate the problem (the need to convert from mass and acceleration to force via f=ma).

So you could save yourself two steps by learning to work with slugs and lbf...

This is probably a good idea, too, since in the aero classes I took, the preferred units were English. You'll need to be proficient in them.

7. Nov 1, 2009

### FredGarvin

I admit it's not the most efficient way of doing things but it can be a good sanity check since it eliminates the confusion with mass units which is easy. If you have the time, work the SI units and then go back and rework the english units until the answers match. You should only have to do it a few times before it sinks in.

8. Nov 2, 2009

### aaronmorg

I talked to my engineering teacher and he said I was going about it the wrong way.

I need, at least, 50% Lift. So I reworked my equation and came out with

FL=$$\frac{(.76764)(.076\frac{lbm}{ft^2})(56.55 in^2)(\frac{47.67\frac{ft}{s}^2}{2})[\frac{ft^2}{144 in^2}]}{[32.2\frac{lbm-ft}{lbf-s^2}]}$$

I come out with .8084 lbf as Lift Force.

With my plane only weighing 27.999g (.0617lbm), is this a reasonable answer? It seems excessive...

9. Nov 2, 2009

### FredGarvin

You did mess up (typing at least) the units for density. Pounds-mass per cubic foot. Also, are you sure about the speed really being 48 ft/sec? 33 mph seems kind of fast for a 28 g aircraft.

10. Nov 3, 2009

### aaronmorg

It seemed right because I calculated 26.14 in-lbf as the energy delivered by the Rubberband.

I converted 26.14 in-lbf to 2.1783 ft-lbf by dividing by 12.

Horizontal Velocity:

VH=$$\sqrt{(\frac{2(2.1783)ft-lbf}{27.999g})[\frac{453.6g}{1lbm}][32.2\frac{lbm-ft}{lbf-s2}]}$$

Comes out to 47.67 ft/s.