Yeah, the same markers that Becka used…. She said it reminded her of bitter dregs, even though he played the Euphonium, not the sousaphone.
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Author: paddlefish
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What is a low interest rate worth?
I’m considering buying a new car. Becka and I have gotten by with one car for ten years, but now with us both working and both jobs too far away to walk to and no good public transit, we’re going to need a second car.
We test drive some Volkswagons this afternoon. I thought I remembered reading that they offered a special 1.2% financing rate. This is a classic sales technique : confuse the consumer. So what, exactly, is 1.2% financing worth?
Loans work over a period of time, at the end of which the principle owed is zero. The amount of the payment owed is a function of the initial value of the loan and the interest rate. If the interest rate is zero, then the payment is simply the initial value divided by the number of payments.
What is interesting about that is that the payment on a loan is a function of two inputs : Initial value and interest rate. Now I do not believe that the car dealer has access to a pool of money at a lower interest rate than everyone else. When they sell you a car, they secretly “contribute” to your loan to make the interest rate lower. They in effect sell you the car for less than you agreed upon — but don’t ever tell you this.
Why should we care? Well, for one thing, you’re paying more taxes than you need to. But the reason I really care is that we might be able to buy our car with either separate financing (through an independent lender) or with cash. In that situation, you really want to know what the present value of the loan at the discounted interest rate is really worth.
This is one situation where Excel is useful. First I studied how interest rate affects payments. For typical car interest rates (between zero and 6 percent), the payment seems to correlate more or less linearly with the interest rate. That’s nice because it means I can make an assumption that simplifies my next calculation. Which is:
=PV((2+E$28)/1200,72,PMT(2/1200,72,$D29))
Where cell E$28 contains the “MSRP”, or price you negotiated. And cell $D29 contains the difference in interest rate between what the car dealer is offering and what current interest rates really are, as quoted by your bank or the internet. Basically I use to financial functions to find out what the present value of a loan with such and such payments over six years is worth today at the real interest rate.
Download a printable version of this table
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I.Fri.
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Wolf eyes.
I took the fancy new “Hubble” lens to the Wolf Center in Forest Lake last weekend. It is the 50-500 Sigma “Bigma” lens. It is not an image stabliized lens.
Nevertheless, I feel I got a few nice pictures with it!
I want to take people up there to see it. There is the Canine Carnival in a few weeks. Also, it is right near the Carlos Avery Wildlife area, which looks like it would be a nice place to take a bike ride.
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Lonely Thursdays
Why is it when I sit in my office alone at work I don’t feel alone? People stop by once in a while, but not that often…. Now that Becka is working at the Textile Center, I am alone in the house when I work at home. I work at home on Thursdays so I can make it to the health sciences orchestra rehearsal at 5:00. Home is a lot closer to the U of M than work. The upside is that I can blast Niel Young at full volume and Becka doesn’t scream bloody murder… Oh well, I’ve got a mocha and a Lucy to keep me company…
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One Fifth as Fast
This weekend Danielle and I rode our bikes along the “Grand Rounds” — a string of parks in Minneapolis that are connected by biking trails. We covered 25 miles in two and a half hours. We took the bike path along the new light-rail track back from Minnehaha Creek Trail up to Washington Ave near the MetroDome. The train passed by a few times, and at one point I noticed the Doppler Effect as I biked past a ringing warning light.
You normally think of the doppler effect as a change in the pitch of a sound when the source of the sound passes you. For example the vrooom sound of the Indy 500, or an ambulance siren. But it also happens when you are the one moving and the source of the sound is stationary.
When the distance between the sound source and the listener is getting smaller, the pitch will sound higher to the listener. When the two are moving apart the pitch will sound lower.
You can train yourself to recognize musical intervals. For example, if you hum the first two pitches of “Somewhere Over the Rainbow”, you have an interval called an “octave”. The first interval of Beethoven’s 5th symphony is a “Major Third”. The first interval of the James Bond Theme “For Your Eyes Only” is a “Major Fifth”. The major third, major fifth and octave correspond to frequency multipliers of approximately 1.25, 1.5 and 2.
If you passed right by a continuous tone on your bicycle, and recognized that the pitch changed by, for example, a Major Third, could you deduce how fast you were traveling?
Suppose the tone was a A 440. That means 440 waves would hit your ear each second. As you approach the tone, you would hear a B 494 and after you passed it you would hear G 392. To alter a pitch from 440 to 494 means that you would travel the distance that 54 A 440 waves occupy in one second (the 54 that you hit while you travel plus the 440 that moved towards you would add up to the 494 that you observed). 54 A 440 waves occupy a distance of (340.29 meters per second divided by 440 per second * 54 = ) 41 meters. 41 meters per second is 91 miles per hour.
| INTERVAL | # HALF STEPS | SPEED |
|---|---|---|
| Diminished Second | 1 | 23 miles per hour |
| Second | 2 | 45 miles per hour |
| Minor Third | 3 | 68 miles per hour |
| Major Third | 4 | 91 miles per hour |
| Fourth | 5 | 113 miles per hour |
| Diminished Fifth | 6 | 136 miles per hour |
| Fifth | 7 | 158 miles per hour |
| Octave | 12 | 272 miles per hour |
Next time you’re watching the Indy 500, see if the doppler effect sound of the racecar engines approaches one octave. Those cars can go as fast as 272 miles an hour at times.
I was waiting to write about this until I could get Becka to help me record some data. I plan to blow a steady pitch on my chickadee okarina while biking down the alley with my laptop recording stuff set up in the driveway. For some reason, Becka hasn’t been too excited to help me with this….
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Tune your glasses with salt?
My friend, Chris, mentioned an interesting scientific observation. We were stting on his front porch for the bimonthly knitting club. I was working on oboe reeds, and everyone else was mostly playing with the dog, Chloe. Chris had bacon and peanut-butter flavored bubbles that he would blow into the air and the dog would chase them. He related a scientific observation he had made earlier.
Chris had been mixing up some salt water in a glass. The spoon occasionally hits the side of the glass, and it makes a ringing sound. At first the pitch of the rining dropped. Then, as the salt dissolved, the pitch of the ringing sound went higher. He said he could tell when it had all dissolved because the pitch stopped going up.
Why does this happen?
The pitch that you hear is a function of resonance inside the glass. Let’s hypothesize that the pitch is due to sound waves going back and forth through the water. The speed of sound in tap water at room temperature is about 1500 meters per second. If we assume that the glass Chris used was about 0.05 meters across, we would expect to have a resonance frequence of about 30000 Hertz.
HUMMM, That is above the normal range of human hearing. I guess that the sound produced by hitting a glass with a spoon is not simply a function of the resonance of the glass. Indeed, if it were, you would expect that an empty glass (one filled with air, in which sound travels at 340.29 meters per second) would have a much lower pitch. In fact, an empty glass has a higher pitch when struck with a spoon.
I believe that what we’re observing is the same phenomena that allows a piano or a string instrument to produce lower pitches on strings of the same (or similar) length to those that produce pitches in higher registers. The low strings are heavier! The relationship shows that frequency (i.e. pitch) is inversely proportional to the density (weight) of the string.
The same rule applies to the ringing of the glass. The water in the glass impedes the motion of the glass in the same way that extra weight impedes the motion of the string. The more water the more weight. Thus the more water weight, the lower the pitch.
I believe the observation that the pitch changes as the salt dissolves is untrue has nothing to do with the salt. While experimenting with salt, water, and a glass just now, I observed that by far the largest change in pitch is due to the shape of the water in the glass when it rings. Stirring the water causes the water to rise up along the sides of the glass, effectively raising the mass of more of the it. This causes the pitch to drop. As you stop stirring as vigorously, the water settles down and the pitch rises. I could not detect any difference in pitch between a glass filled with tap water, and one filled to the same level with salt water.
Sorry to hit you all with this again, but I have updated a few calculations and made some interesting new observations.
A friend sent me a link to a website with an interesting new material called liquid metal. They have a video of a steel ball bearing bouncing on a plate of this metal, which shows the material’s surprising elasticity. The steel ball bounces for much longer than on a similar plate of titanium or steel.
The bouncing sound was somewhat like a metronome. Except that I noticed that the time it took for the ball to bounce got shorter the longer it bounced. So it starts out going ping — ping — ping — ping quite regularly but after a while it is faster, ping — ping — ping — and eventually is bouncing so quickly it sounds like a buzzing noise.
The ball bearing appears to be behaving as an oscillating spring with a decay. But what kind of decay? Was curious.
So I found an application called Wire Tap Pro” which has a 30 day free trial. I recorded the audio track of the bouncing ball bearing and opened it in sound studio.
I took a few data points from the audio track. I selected an interval of ten bounces, and sampled about every five seconds through the 40 second recording. I added one last point at the end at the best resolution I could discern (it got hard to read because the bounces got quieter and closer together).
I typed the data into a text file using my favorite text editor, vim.
1 start end 2 0.134 2.635 3 4.969 7.146 4 9.956 11.813 5 15.180 16.695 6 19.969 21.211 7 25.043 25.957 8 29.986 30.613 9 34.982 35.291 10 36.424 36.637
What I really wanted to plot was the time since ball was released on the X axis, and duration of the bounce on the Y. So I needed to plot the average of the start and end time versus the average of the duration. I used perl, my digital swiss army knife, to convert this file.
1 perl -pi.bak -e 's[([\d.]+)\t([\d.]+)][($1+$2)/2.0 . "\t" . ($2 - $1)/10.0]e;s/start/time/;s/end/duration/;' bounce.dat
This resulted in converted data:
1 1.3845 0.2501 2 6.0575 0.2177 3 10.8845 0.1857 4 15.9375 0.1515 5 20.59 0.1242 6 25.5 0.0914000000000001 7 30.2995 0.0626999999999999 8 35.1365 0.0308999999999998 9 36.5305 0.0213000000000001
This I could plot using GNUPlot and this plot file:
set term postscript eps color 48;
set output "test.eps";
set size 3.2,5.2;
set origin 0,0;
set multiplot;
set size 3,5;
set origin 0.1,0.1;
set pointsize 20;
set xlabel "time since release (s)";
set ylabel "duration (s)";
f(x) = a*x + b
fit f(x) '/Users/arahn/bounce.dat' via a,b
set xtic rotate by -45;
set title "Bouncing Of a Steel Ball Bearing";
plot f(x),'/Users/arahn/bounce.dat' with points pt 7 ps 3 ti "liquid metal";
To get this pretty graph:
gnuplot also gave me the regression values, intercept of 0.256846 s and slope of -0.00645081 s/s. This means that each bounce lasts 99.354919% as long as the bounces in the previous second. I was amazed at how well the data fit the regression line!
I would expect that decay would act on the total energy of the system — but we’ve shown that it correlated with the duration of the bounce. However, the tricky part is that you can’t really draw any conclusion based on how the duration of the bounce correlates with how long ago the ball was released. Energy is lost when the ball strikes the ground. As time goes on, it is striking the plate more and more often. What I’d really like to do is correlate the how many times it has hit the ground with how much energy it has. My hypothesis is that it would be an exponential decay in energy with respect to each bounce.
To determine how many times the ball has hit the ground, we could count them all. Or we could use calculus. The integral of the rate (clicks per second) is a function determining how many clicks have passed at time t.
click duration = d = mx + b click rate = 1 / d = 1 / (mx + b) num clicks = N = integral ( 1/ (mx + b) ) evaluated from zero to t N = ( ln( mt + b ) - ln( b ) ) / m
We can turn that around to find out how the duration of a click is related to the number of times the ball has bounced:
m N = ln( d ) - ln( b ) d = b em N
I went back to see if this new data jives. We expect the first click (index zero) to have a duration of b, or 0.257 seconds. The first bounce in the audio recording is 0.261 seconds. Here is a table comparing the actual bounce times versus expected:
1 hit_num actual_time computed_time 2 0 0.261 0.257 3 10 0.242 0.242 4 20 0.228 0.227 5 30 0.212 0.213 6 40 0.197 0.199
Outstanding! The numbers are almost identical. Before you berate me for whimping out after 40, keep in mind what a nuisance it is to try and count all the blips in the recording.
Let see, we also know that the click duration is equal to the velocity (in meters per second) divided by the acceleration of gravity (G). Thus
velocity = b G em N
Here my hypothesis is wrong : Energy is proportional to the square of velocity. But we’ve shown that the duration of each bounce is proportional to velocity, and therefore the square root of energy. This reveals something interesting about inelastic collisions that I didn’t realize before.
An experiment for next time: We’ve shown that the amount of energy lost each time the ball strikes the plate is proportional to how quickly the ball strikes the plate. I wonder if the amount of noise that it makes is proportional to either the speed, or the energy with which it strikes? We could use the same sound recording to look at the size of the sound wave and relate it to how closely together the clicks occur.
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Travel
I sometimes wish I could travel back in time to when I was a child in the garden with a pumpkin.
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Geeky Oboe Guy
I just wanted to share some oboe info. One thing was this: I just have to brag about my new oboe reed guillotine invention. You can buy one of these Reiger Tools for $150. Or you can spend a few hours with a router, drill, etc…
Also, the other day I bought some english horn reed supplies. English horn reeds are pretty similar to oboe reeds, but the guy at the store (Wes, who plays with the Metropolitan Symphony) gave me some info on the differences. I recopied it on a napkin during knitting yesterday:
I forgot to show on the diagram that the bottom of the tip begins at 51.5mm and probably goes up about a mm to 52.5. That leaves 3.5 mm or so for the tip. He also indicated that he prefers to rough in the back first, then let it dry over night. Then the next day work on the tip *dry* (is this crazy?) and then wet it, clip it open at 57mm, and tie on the wire. I’ve never consciously made a “front” and a “back” to my reeds before, but he said since the lower lip provides more cushion that the window should therefore be bigger (46mm instead of 45mm).
I’ll try my hand at English Horn reeds sometime soon, I’m sure. As if I didn’t have enough to do with my oboe and Brahms’ 4th, hee hee hee.