RE: Badge Placement...

["FlyinVR6" <FlyinVR6@li.net>]

All of us better get our boots on because the bullshit is getting thick!
-FlyinVR6

-----Original Message-----
From: owner-jettaglx@igtc.com [mailto:owner-jettaglx@igtc.com]On Behalf
Of Ng, Kevin (NationsCredit)
Sent: Friday, August 06, 1999 6:56 AM
To: 'jettaglx@igtc.com'
Subject: RE: Badge Placement...


Tsk, tsk. Khan, you may have aced your math class, but you obviously have
some more studying to do in your physics class. Your math assumes that we
live in a perfect world, in a vacuum, devoid of atmospheric conditions and
the effects of gravity. For instance, in order for your method to work, the
laser attached to a tall building or crane would have to be *fixed* relative
to the car. The accuracy of the laser is reduced the farther away the laser
is, because in the real world, high winds, the heat of the sun, and
modulations in the earth's crust would significantly alter the laser's
position relative to the car, thus rendering your method useless. The car's
position is never fixed either. By the time the experiment can be setup
correctly, air pressure in the tires would have decreased, and the springs
would have sagged, thereby changing the another key variable. I don't even
want to get into the effects of gravity.

Remember, the key law of motion here is that the angle of the dangle is
directly proportional to the beat of the meat.

Kevin Ng, Ph. D. V.W.
97 GLX Windsor Blue

> -----Original Message-----
> From:	Khan Klatt [SMTP:kmk@pacificrim.net]
> Sent:	Thursday, August 05, 1999 9:51 PM
> To:	jettaglx@igtc.com
> Subject:	Badge Placement...
>
> 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