Badge Placement...

[Khan Klatt <kmk@pacificrim.net>]

John wrote:
>Anyone have tips for badge placement?  I'm going to replace my Jetta
>badge with a Vento badge, and I'm unsure of my ability to get it on
>level. :-)

Well, other than using a gravitationally aligned laser using weights 
and pendulums on really long strings (you know sin theta = zero as 
theta goes to zero).... you could just use a dry erase pen with a 
fine tip to mark under the existing badge. Then, work off the badge 
with floss, and then affix the new badge above the line. Disclaimer: 
try the dry erase pen an an inconspicuous location before you go 
trying it on your trunk lid.

If you insist on using the sin theta -> 0 mechanism, this is what I suggest.

Find a really really tall building, or perhaps use a crane from one 
of these construction sites in your neighborhood. Tie four strings to 
the top of the crane, long enough to reach the ground. The longer, 
the better. I'd say that a 20 yard length is probably close enough to 
provide a straight enough line. For the record, 20 yards is about 60 
feet is about 720 inches.

At this stage, picture that the left bottom edge of your badge is one 
point in a triangle, and the right bottom edge is another point in 
the triangle, and the point where the laser is suspended is the third 
point in the triangle. This means your triangle is 720 units by 720 
units by 2 units. The angle created at the top of the crane as the 
laser swings on the strings from one corner to the other we'll call 
theta. The angle can be calculated by taking the opposite side of the 
triangle, and dividing by the hypotenuse, and taking the inverse sin 
of the number that that produces. So, the opposite would be 2, and 
the hypotenuse would be 720. Or, we'd want the inverse sin of 
0.0027778, which roughly translates into an angle of 0.1591 degrees.

We can then calculate the length of the arc that is created by 
swinging the laser in an arc of 0.1591 degrees by the equation l= 
theta r. In the general case, theta would be 2pi, and therefore the 
circumference of the whole circle would be 2*pi*r. Anyhow, back to 
the calculation. First, we describe the angle 0.1591 in radians, 
which happens to be 0.27767. So, 0.27767 * 720 = 1.9992506.

So, again, if the laser is swung along a radius of 720 inches, and 
the width of the badge is 2 inches, the arc that is swung out by the 
laser is 1.9992506 inches.

In other words, the difference in length by swinging it in the arc 
vs. the straight line is 2 - 1.9992506, or 0.0007494 inches. In other 
words, the difference is less than 8/10000th of an inch. Effectively, 
this proves that you come as close to a straight line as is 
practically necessary for the purpose of affixing your badge to your 
car using this mechanism.

So, place your badge on the back of the car, swing the laser in a 
small arc of less than two inches, and make it so that the bottom of 
the badge aligns with the effectively straight line that the laser 
light makes on your trunk. If you don't need 8 ten thousands of an 
inch in accuracy, you can always just suspend the laser from a ten 
foot height or something more reasonable. You can calculate that 
scenario, just use 2 inches as the opposite, and 120 inches as your 
hypotenuse.

It is left as an exercise to the reader to provide the equation that 
shows the percentage of accuracy of the curve to the straight line as 
a function of the radius. (as the radius -> infinity, the line gets 
closer and closer to straight).

And when I was in math class, people used to complain that they'd 
never use this stuff in the real world. Tsk, tsk.

-Khan