Bear with me.  It's going to take a bit of prelude to get to the point of this
article.  If you already understand all there is to know about angle notation,

The ancient Babylonians counted using a base sixty system.  Unfortunately,
this system has survived to today in the numerics we use to count time and
angle. We write time as h:m:s where sixty seconds (s) = one minute (m) and
sixty minutes = one hour (h).  Similarly, angles are written as d:m:s with the
same relationships (60 arcseconds = 1 arcminute, 60 arcminutes = 1 degree).
Mathematicians normally add the prefix 'arc' to distinguish the fact that
they're talking about angle and not time.  (There really ought to be a special
circle in hell for anyone who uses the same term for two completely disparate
units or, like the American gallon, redefines an existing unit.)

For most practical applications it's much more convenient to express angles as
decimal numbers.  This raises the problem of converting between the two
notations.

Going from d:m:s to decimal notation is straightforward.  Consider converting
12:34:56 (12 degrees, 34 arcmin, 56 arcsec) to decimal degrees.  We know that
34 arcmin is 34/60 of a degree.  We also know that there are 60*60 = 3600
arcsec in a degree.  So the 56 arcsec is 56/3600 degrees.  Adding them, we
have:

12 deg + 34/60 deg + 56/3600 deg = 12.582221... degrees

or, in general form:

d:m:s = d + m/60 + s/3600 decimal degrees

If your print calls out 12:34:56 d:m:s and you need the tangent of that angle
you'll need to perform the above calculation to get the decimal degrees
to feed to the tangent function.  (Better scientific calculators have this
conversion built-in but the less expensive ones often lack it.)

Converting from decimal to d:m:s isn't very difficult.  Using 12.582221
decimal degrees as an example:

Extract the integer degrees:

12.582221 = 12 + 0.58222112 degrees

Multiply the remainder by 60 (arcmin/deg):

60 * 0.582221 = 34.93326

Extract the integer arcmins:

34.93326 = 34 + 0.9332634 arcminutes

Multiply the remainder by 60 (arcsec/arcmin):

60 * 0.93326 = 55.9956~56 arcseconds

Again, better scientific calculators have a single key to do this conversion.
However, if yours lacks it, no worry.  You won't be doing it frequently and
the procedure above is straightforward.

Most scientific calculators can deal with angles in decimal degree notation,

What the hell are radians and why do we need them?  Isn't d:m:s notation
confusing enough?  Now you're telling me that we need two more ways of
expressing angles?

When doing mathematics, it's much more useful to express angles in a
notation such that the angle so expressed, when multiplied by the radius of a
circle, yields the length of the arc on the circle subtended by that angle.

Consider a 90 deg angle.  It subtends one-quarter of the circumference of a
circle or an arc length of 2*pi*r/4 (2*pi*r = the circumference of a circle
whose radius is 'r').  We want this angle (we'll call it 'A') expressed in

A (rad) * r = 2 * pi * r / 4

That is, the angle in this radian notation, multiplied by the radius of the
circle, equals the length of the arc on said circle subtended by this angle.

Cancelling the 'r's, we have:

Since we assumed that A=90 deg, we now have a relationship between degrees and

or:
or:
1 radian = 180/pi degrees =~ 57.295831 deg

Which makes things pretty simple.  If we have degrees and want radians,
multiply degrees by 180/pi.  If we have radians and want degrees, multiply
radians by pi/180.  Rather than trying to memorize that, simply remember that
a full rotation, 360 deg, equals 2*pi radians.

-------------------------------

The French are never happy with any measurement system they didn't personally
invent.  They thought that 90 degrees was an awkward number for a right angle
so they 'metricized' it to be 100 grads.  I don't remember the details but
their argument for this aberration revolved around the fact that slopes
expressed in percent (as we express the slope of hills in road-building) would
then convert directly to grads without the need to do any calculation.
Apparently the French are as intimidated by arithmetic as they are by warfare.

Don't worry about grads.  They're only used by the French and a few other
equally mis-guided Europeans.  In 30+ years of doing mathematics for a living,
I *never* had to convert any angles to grads.  Should you ever need to do so,
the relationships are:

-------------------------------

Back to radians and the mathematicians.  If you ask a mathematician to
calculate the sine of an angle, he'll write down something like this:

sin (x) = x - x^3 / 6 + x^5 / 120 - ...

This is the 'series expansion' for the sine of x and it's only a valid
equation if x is expressed in *radians*.  By using enough terms in this series,
you can calculate the sine to whatever precision you desire.  In fact, early
calculators used a series similar to this to calculate trig functions.
(Today, with cheaper memory, they use a table lookup scheme.)

Now, if we look at this equation, we can see that, if x is a small number
(i.e., a lot less than one), x^3/6 is a lot less than x and x^5/120 is a lot
less still.  In other words:

sin (x) ~= x for x << 1

A numerical example will verify this.  Using my calculator:

sin (5 deg) = 0.087155742

5 deg * (pi/180) rad/deg = 0.087266462

In other words, 5 deg expressed in radians is pretty close to the sine of 5
deg.  Some other examples:

angle = 1 deg: sine = 0.017452, radians = 0.017453, error = 0.005077 %
angle = 2 deg: sine = 0.034899, radians = 0.034907, error = 0.020311 %
angle = 3 deg: sine = 0.052336, radians = 0.052360, error = 0.045707 %
angle = 4 deg: sine = 0.069756, radians = 0.069813, error = 0.081278 %
angle = 5 deg: sine = 0.087156, radians = 0.087266, error = 0.127037 %
angle = 6 deg: sine = 0.104528, radians = 0.104720, error = 0.183005 %
angle = 7 deg: sine = 0.121869, radians = 0.122173, error = 0.249205 %
angle = 8 deg: sine = 0.139173, radians = 0.139626, error = 0.325666 %
angle = 9 deg: sine = 0.156434, radians = 0.157080, error = 0.412420 %
angle = 10 deg: sine = 0.173648, radians = 0.174533, error = 0.509506 %

-----------------------
Another aside.  When physicists write the differential equation for a swinging
pendulum, they assume small angular motions of the pendulum and use this
sin(x) ~= x approximation.  This allows them to obtain a simple equation for
the period of the pendulum.  The terms of the series they throw away account
for an error that clock builders call the 'circular error' which will cause
timekeeping errors if the pendulum is allowed to swing through too wide an
arc.  Ask yourself:  Have you ever seen a pendulum clock where the pendulum
swings through a wide arc?
-----------------------

So, what good is all this to us as metalworkers?  I recently bought a cheap
(\$10) laser level at Harbor Freight.  The vials on it are adequate for
carpentry work but hardly anything one might use to align machinery.
Nevertheless, this 'instrument' consists of a flat base to which is attached a
laser capable of projecting a tiny dot or line over a considerable distance.
The light always travels in a straight line (Einstein will argue with that but
that's not anything we need to worry about here).  In other words, this thing
is a very accurate, very long sine bar even if you never bother to look at the
bubbles in the vials.

I laid this 15" long level across the top of my 12" long surface plate such
that the projected laser line hit a vertical 6" scale on a bench 15' away on
the other side of the garage.  I noted the reading on the scale.  Then I
placed a 0.004" feeler gage under one end of the level.  The reading on the
scale changed by 1/16".

Now for some arithmetic.  Putting the feeler gage under the level turned it
into a sine bar.  My 'stack height' is 0.004" and the length of the sine bar
is 12".  What angle does that correspond to?  We have, from trig:

sin x = 0.004/12

But the angle 'x' is tiny so sin x ~= x.  So:

Does this make sense?  Well, the end of the laser beam is swinging around
a circle with a radius of 15' = 180".  Our angle is already in radians so the
arc length at that radius is:

333e-6 rad * 180" = 0.06"

which is pretty close to the 1/16" I measured with my bifocals while fighting
the laser speckle.  (Ever wonder what causes that speckle effect in laser
light?)

change of 333 urad with a \$10 Chicom tool and a ruler!  The Starrett
Machinist's Level is advertised to resolve (same as detect) 90 arcsec.  An
arsec is about five urad (check that for yourself) so that's a resolution of
450 urad.  I did better than that with a double sawbuck's worth of gear and
some math.  The Starrett level costs \$280.  (The Starrett Master Precision
Level claims 10 arcsec (50 urad) resolution, but it costs \$520).

The longer the path traveled by the beam, the more the deflection on the scale.
I put a first surface mirror where the scale was and moved the scale back on to
the surface plate.  Now the light has to travel ~30' to reach the scale.  Not
unsurprisingly, the 0.004" gage now moves the beam through about 1/8".  So, by
bouncing the beam over a longer distance we can make detecting a small angle a
lot easier.  Unfortunately, even with a quality first surface mirror, the laser
spot is smeared to a circle about 0.5" diameter by the time it reaches the
scale.  This makes reading its location tricky.  I picked out a particular
feature on the distorted spot and read its location on the scale.  (I think I
have an idea about how to deal with this problem - see below.)

We all know that lathes don't really need to be level.  Lathes produce
precision parts on ships that roll. (Lathes on submarines were sometimes
mounted with the bed vertical.)  We're really using the level to detect any
twist in the lathe bed.  What we're doing is trying to get one end ot the bed
to point the same direction in space as the other end.  The laser level with a
long enough light path will do this with a resolution that approaches or
exceeds what you can do with an expensive level that you can't afford and will
use very seldom.

Some practical remarks here.  The laser level packaging claims that it will
project its beam 1500'.  Those must be 'Sears' feet. <g>  Perhaps the
hyper-sensitive eye can detect the presence of the beam 1500' away but don't
count on doing any useful measurement at that distance.  Projecting the beam
out of doors to get greater distance isn't going to work during daylight.  The
red laser beam is washed out by sunlight.  Better is to use mirrors to bounce
the beam around a darkened shop.  You'll want to use first surface mirrors.
As the name suggests, these are mirrors where the reflective coating is on the
front of the mirror rather than the back, as most cheap mirrors have.  If you
use a cheap mirror, the triple refraction (at the glass surface and at the
mirroring) will muddy the image and make it harder to read.  I have some first
surface mirrors that I'll be happy to lend to anyone who wants to try this.

I found that, at 15', reading the beam location to better than 1/16" is
difficult.  I have an idea, as yet untried, on how to deal with that.  If I
build a simple circuit consisting of little more than a biased
photo-transistor powered by a 9v battery, I can use it as a light amplitude
detector.  (The voltage across the biasing resistor will be proportional to
the light intensity falling on the photo-transistor.)  If I attach this to a
height gage, I'll be able to measure the height of the maximum signal coming
from the laser light.  This will take my bifocals and the laser speckle
problem out of the equation.

Now get out there and think of some other neat things one can do with a thirty