I mentioned this in class late last week:
The mathematical technique of a Fourier Transform (see pg 230 of the Wolfson physics textbook we use, but also any number of math or physics textbooks) has been used to solve the mystery of the unique opening chord of the Beatles' "Hard Days Night". The mathematical software used to do this is capable of isolating the fundamental and all harmonics of a note played by a particular instrument, which made it possible to find the missing fourth instrument (George Martin on piano) responsible for the sound.
The article in Wired includes a sample of this chord, and a link to the pdf version of the report.
Showing posts with label physics. Show all posts
Showing posts with label physics. Show all posts
Monday, November 10, 2008
Tuesday, May 27, 2008
Calculus and Physics
I Will Derive!
Doug sent around the URL for this YouTube video but didn't get around to embedding it in a post here. (My brother also sent it to me, from a different blog, so this has definitely gone viral in some part of the blogosphere.) If you click on the image to go to the YouTube site, you can read all of the lyrics in the "more info" area.
Obligatory math/physics comment:
This is a nice example of the application of calculus to motion in physics, a topic taken up in chapter 3 of the calculus textbook and chapter 2 of the physics textbook.
Doug sent around the URL for this YouTube video but didn't get around to embedding it in a post here. (My brother also sent it to me, from a different blog, so this has definitely gone viral in some part of the blogosphere.) If you click on the image to go to the YouTube site, you can read all of the lyrics in the "more info" area.
Obligatory math/physics comment:
This is a nice example of the application of calculus to motion in physics, a topic taken up in chapter 3 of the calculus textbook and chapter 2 of the physics textbook.
Sunday, February 24, 2008
Compressing a Quarter
After compared to Before ...
A standard quarter is electromagnetically compressed to the size of a dime in, literally, the blink of an eye, a flash of light, and a loud crack.
The forces on the coin are produced by an estimated 100 kA current in a 10-turn copper solenoid winding. The induced current (about 1 MA) in the outer edge of the copper core of the quarter interacts with an estimated 58 T magnetic field produced by the solenoid, compressing the quarter radially. The peak field last for about a millionth of a second because the coil is vaporized by that current.
The energy needed (about 4600 J) is delivered by a 178 uF capacitor charged to 7200 V. The capacitor, which contains about 1.28 coulombs of charge, takes several minutes to reach its final voltage. A display next to the demonstration included a page on the principles of operation of a quarter shrinker from "Stoneridge Engineering", the Teslamania web site of Bert Hickman that gave additional details for a system similar to the one we saw. My independent calculations at the bottom of this article roughly confirm what is on that page and the display at the magnet lab.
Side comment. You would get 4600 J of energy if you dropped 47 kg (about 100 pounds) a distance of 10 m (about 33 feet). Now imagine all of that energy concentrated int the small area of a quarter. Splat.
Tools of the trade:
An assembled coil is in the background. The coil itself (left foreground) is about 10 turns of 14 gauge wire (wire that would normally carry no more than 15 A in commercial use, and will melt if used with 166 A). The quarter is sandwiched between two cylinders of G10 fiberglass (right background) to center it within the coil. (Wood had been used in the past, but the G10 survives and can be reused from year to year.) Tape holds it all in place.
Centering the quarter and ensuring it is perpendicular to the magnetic field is crucial to making sure the forces compress the coin rather than twist it. Oh, yes, and the vial contains fragments of copper coils used in past experiments. The lighter colored pieces are probably stainless steel chipped off of the box that is used to contain the explosion of the coil.
This last photo shows the interior of the box used to contain the explosion of the magnet coil. You can see the copper from the used coils embedded in it. I also notice that the stainless steel panels on those three sides appear to have been added after the fact to the inside of Lexan panels that appear to have been the original design plan.
The Physics Details:
I collected a lot more details this year than I managed to get last year, correcting some of the information I used for back-of-the-envelope example in PHY2049 this year and last. [I had remembered the 1 MA current, but not that it was the induced current in the coin or the number of turns, and did not have enough info to estimate the R or L of the circuit being used here.] I now know he used 14 AWG 200C copper magnet wire with 10 turns in the coil. That info, and my estimate that the coil has an inner diameter of 2.4 cm (to fit around a quarter) and a length of between 1.7 cm (absolute minimum for the wire diameter) and 2 cm, are essential to a qualitative understanding of what is going on.
The length of the coil plus 10 cm for each lead is about 1 m, so we can estimate its resistance at about 0.008 ohm. All other conductors are large bus bars and will contribute little to the resistance of the circuit.
The coil that produces the magnetic field is too short to be correctly modeled as a solenoid. If we do that anyway, however, we require about 92 kA to produce 58 T, while 65 kA (see below) produces 41 T. These numbers assume 10 turns and a 2 cm length. Only 78 kA is needed to make 58 T (or 48 T from 65 kA) if the solenoid is 1.7 cm long. We always want the coil as tightly wound as possible!
We also get an inductance L = 2 uH for a 2 cm coil. An on-line calculator (of unknown reliability) says those dimensions would give 1.8 uH, while 1.7 cm gives 2.0 uH. I will use L = 2 uH as a conservative value. Notice that a shorter length makes the inductance bigger, which is a bad thing.
The RC time constant of 1.4 us (micro seconds), with an initial current of 900 kA, tells us what would happen if there was no coil (inductance) in the circuit. Unfortunately, the coil does not like a rapidly rising current. It would take 250 us (0.25 ms) for the current to reach 570 kA if we had just the coil with an ideal 7200 V battery. However, this is actually an un-driven LCR circuit, with all three elements playing a role. The circuit has an inductive time constant of 500 us and a natural frequency of about 53,000 rad/s. The solution to this problem says a peak current of 65 kA will be reached after 19 us. A smaller inductance makes the current bigger, by the way. The estimated inductance is a critical quantity. Reducing the inductance to 1.8 uH will increase the current to 78 kA. Notice that this is just what we need to produce 58 T in a 1.7 cm coil.
Faraday's Law says the large dB/dt produces a large EMF around the edge of the coin, which acts like a single turn in a 10:1 transformer. This leads to a rough estimate of an induced current of 650 kA (perhaps 780 or 920 kA) around the edge of the coin.
Even with "only" I = 65 kA and B = 41 T, the compressive force due to the 650 kA current induced in the outer edge of the coin would be something like 2 MN. If we really have I = 78 kA and B = 58 T, the force increases to 3.4 MN.
This force only acts for a micro second or so at the peak of the sine function that describes the current (before the coil is torn apart by the equal and opposite repulsive force of the field on the coil). The total impulse is not very large, although the total energy is significant even if a lot of it is wasted.
A standard quarter is electromagnetically compressed to the size of a dime in, literally, the blink of an eye, a flash of light, and a loud crack.
The forces on the coin are produced by an estimated 100 kA current in a 10-turn copper solenoid winding. The induced current (about 1 MA) in the outer edge of the copper core of the quarter interacts with an estimated 58 T magnetic field produced by the solenoid, compressing the quarter radially. The peak field last for about a millionth of a second because the coil is vaporized by that current.
The energy needed (about 4600 J) is delivered by a 178 uF capacitor charged to 7200 V. The capacitor, which contains about 1.28 coulombs of charge, takes several minutes to reach its final voltage. A display next to the demonstration included a page on the principles of operation of a quarter shrinker from "Stoneridge Engineering", the Teslamania web site of Bert Hickman that gave additional details for a system similar to the one we saw. My independent calculations at the bottom of this article roughly confirm what is on that page and the display at the magnet lab.
Side comment. You would get 4600 J of energy if you dropped 47 kg (about 100 pounds) a distance of 10 m (about 33 feet). Now imagine all of that energy concentrated int the small area of a quarter. Splat.
Tools of the trade:
An assembled coil is in the background. The coil itself (left foreground) is about 10 turns of 14 gauge wire (wire that would normally carry no more than 15 A in commercial use, and will melt if used with 166 A). The quarter is sandwiched between two cylinders of G10 fiberglass (right background) to center it within the coil. (Wood had been used in the past, but the G10 survives and can be reused from year to year.) Tape holds it all in place.
Centering the quarter and ensuring it is perpendicular to the magnetic field is crucial to making sure the forces compress the coin rather than twist it. Oh, yes, and the vial contains fragments of copper coils used in past experiments. The lighter colored pieces are probably stainless steel chipped off of the box that is used to contain the explosion of the coil.
This last photo shows the interior of the box used to contain the explosion of the magnet coil. You can see the copper from the used coils embedded in it. I also notice that the stainless steel panels on those three sides appear to have been added after the fact to the inside of Lexan panels that appear to have been the original design plan.
The Physics Details:
I collected a lot more details this year than I managed to get last year, correcting some of the information I used for back-of-the-envelope example in PHY2049 this year and last. [I had remembered the 1 MA current, but not that it was the induced current in the coin or the number of turns, and did not have enough info to estimate the R or L of the circuit being used here.] I now know he used 14 AWG 200C copper magnet wire with 10 turns in the coil. That info, and my estimate that the coil has an inner diameter of 2.4 cm (to fit around a quarter) and a length of between 1.7 cm (absolute minimum for the wire diameter) and 2 cm, are essential to a qualitative understanding of what is going on.
The length of the coil plus 10 cm for each lead is about 1 m, so we can estimate its resistance at about 0.008 ohm. All other conductors are large bus bars and will contribute little to the resistance of the circuit.
The coil that produces the magnetic field is too short to be correctly modeled as a solenoid. If we do that anyway, however, we require about 92 kA to produce 58 T, while 65 kA (see below) produces 41 T. These numbers assume 10 turns and a 2 cm length. Only 78 kA is needed to make 58 T (or 48 T from 65 kA) if the solenoid is 1.7 cm long. We always want the coil as tightly wound as possible!
We also get an inductance L = 2 uH for a 2 cm coil. An on-line calculator (of unknown reliability) says those dimensions would give 1.8 uH, while 1.7 cm gives 2.0 uH. I will use L = 2 uH as a conservative value. Notice that a shorter length makes the inductance bigger, which is a bad thing.
The RC time constant of 1.4 us (micro seconds), with an initial current of 900 kA, tells us what would happen if there was no coil (inductance) in the circuit. Unfortunately, the coil does not like a rapidly rising current. It would take 250 us (0.25 ms) for the current to reach 570 kA if we had just the coil with an ideal 7200 V battery. However, this is actually an un-driven LCR circuit, with all three elements playing a role. The circuit has an inductive time constant of 500 us and a natural frequency of about 53,000 rad/s. The solution to this problem says a peak current of 65 kA will be reached after 19 us. A smaller inductance makes the current bigger, by the way. The estimated inductance is a critical quantity. Reducing the inductance to 1.8 uH will increase the current to 78 kA. Notice that this is just what we need to produce 58 T in a 1.7 cm coil.
Faraday's Law says the large dB/dt produces a large EMF around the edge of the coin, which acts like a single turn in a 10:1 transformer. This leads to a rough estimate of an induced current of 650 kA (perhaps 780 or 920 kA) around the edge of the coin.
Even with "only" I = 65 kA and B = 41 T, the compressive force due to the 650 kA current induced in the outer edge of the coin would be something like 2 MN. If we really have I = 78 kA and B = 58 T, the force increases to 3.4 MN.
This force only acts for a micro second or so at the peak of the sine function that describes the current (before the coil is torn apart by the equal and opposite repulsive force of the field on the coil). The total impulse is not very large, although the total energy is significant even if a lot of it is wasted.
Tesla at the Magnet Lab
"Conducting" Electricity
The large Tesla coil shown here was producing about 250 kV at 200 kHz. The presence of an aluminum rod nearby alters the spark pattern because the current being carried off by the sparks is drawn to the grounded conductor rather than something like the overhead lights up by the ceiling.
The sparks look purple because of the emission lines from nitrogen gas. (The spark results when air molecules, mostly nitrogen, get ripped apart by the high voltage. The current is carried by a plasma.) A bare hand works almost as well as an aluminum rod, since your body is a conductor.
At high frequencies (such as 200,000 Hz), the current flows over the surface of the body. The only risk is from a burn at the point of contact. Based on the reaction of one other demonstrator, you can definitely feel it. This photo was really spectacular. The yellow spots you see are where it is burning his hand.
Important Note:
Frequency is the difference between life and death. A low frequency (such as 50 or 60 Hz) Alternating Current would be fatal at those voltages. That would go through your body and, partly because it is so well matched to the frequency our nervous system operates at, would stop the heart. In contrast, we easily get 200 kV from a van de Graaff generator in lecture demos, but the resulting small Direct Current only produces a painful shock.
Musical Reindeer
The sparks emanating from a small stuffed reindeer sitting on a smaller Tesla coil produced music. (Click the photo for a much bigger version.)
What was happening was that the air gets heated by the glowing discharge you see in the photo. If the intensity of the sparks varies at 5000 times per second (too fast to see or photograph), you get a sound wave of 5000 Hz produced in the air instead of the crackling sound of the sparks. They modulating the amplitude of the voltage delivered to the coil (and putting wires in the stuffed animal to produce many sharp points for spark production) with a music source, and you hear music.
The music went away, or was altered, if a rod was used to draw off the current in a single spark rather than the discharge you see here. Very nice.
The large Tesla coil shown here was producing about 250 kV at 200 kHz. The presence of an aluminum rod nearby alters the spark pattern because the current being carried off by the sparks is drawn to the grounded conductor rather than something like the overhead lights up by the ceiling.
The sparks look purple because of the emission lines from nitrogen gas. (The spark results when air molecules, mostly nitrogen, get ripped apart by the high voltage. The current is carried by a plasma.) A bare hand works almost as well as an aluminum rod, since your body is a conductor.
At high frequencies (such as 200,000 Hz), the current flows over the surface of the body. The only risk is from a burn at the point of contact. Based on the reaction of one other demonstrator, you can definitely feel it. This photo was really spectacular. The yellow spots you see are where it is burning his hand.
Important Note:
Frequency is the difference between life and death. A low frequency (such as 50 or 60 Hz) Alternating Current would be fatal at those voltages. That would go through your body and, partly because it is so well matched to the frequency our nervous system operates at, would stop the heart. In contrast, we easily get 200 kV from a van de Graaff generator in lecture demos, but the resulting small Direct Current only produces a painful shock.
Musical Reindeer
The sparks emanating from a small stuffed reindeer sitting on a smaller Tesla coil produced music. (Click the photo for a much bigger version.)
What was happening was that the air gets heated by the glowing discharge you see in the photo. If the intensity of the sparks varies at 5000 times per second (too fast to see or photograph), you get a sound wave of 5000 Hz produced in the air instead of the crackling sound of the sparks. They modulating the amplitude of the voltage delivered to the coil (and putting wires in the stuffed animal to produce many sharp points for spark production) with a music source, and you hear music.
The music went away, or was altered, if a rod was used to draw off the current in a single spark rather than the discharge you see here. Very nice.
Monday, January 21, 2008
Cause of Minnesota Bridge Collapse
My dad may be retired but he is still a member of ASCE, the professional society for civil engineers. He forwarded a copy of the urgent e-mail sent by ASCE to all members about the cause of the collapse of the I-35W bridge in Minnesota last fall that killed 13 people.
The collapse was due to a design error, not corrosion or maintenance problems that were assumed to be the problem during the discussions immediately after the disaster.
You can read the NTSB Safety Recommendation Letter and a detailed interim report by two engineers with the Federal Highway Administration on one specific part of that bridge design on the web. Links to these reports are also prominently featured on the ASCE web site.
It is unclear if the undersized plates used at two key joints were the result of a calculation error, an error in developing the plans, or even a drafting error. The documentation on file is still unclear at this point. What is clear is that you don't need a computer to do the calculation, one part of which would be a basic PHY2048 problem. What is important is that, wherever the error was made, the fact that this error ended up in the final plans indicates a significant error in the design process and a failure of the review process used by the engineers responsible for the design. The design review process itself, particularly when renovations are made to an existing bridge, has become the focus of the investigation.
Speaking of presentations, I'd love to hear one of our alumni walk through the part of the calculation I know nothing about, the stress calculation and the treatment of rivets in the net force and torque calculation.
The most striking thing to me was the similarity between the way they presented their results and the way the West Point Bridge Design computer program works to highlight strong and weak points in a design, although this particular design element is not part of the simple bridge designs used in that program.
The collapse was due to a design error, not corrosion or maintenance problems that were assumed to be the problem during the discussions immediately after the disaster.
You can read the NTSB Safety Recommendation Letter and a detailed interim report by two engineers with the Federal Highway Administration on one specific part of that bridge design on the web. Links to these reports are also prominently featured on the ASCE web site.
It is unclear if the undersized plates used at two key joints were the result of a calculation error, an error in developing the plans, or even a drafting error. The documentation on file is still unclear at this point. What is clear is that you don't need a computer to do the calculation, one part of which would be a basic PHY2048 problem. What is important is that, wherever the error was made, the fact that this error ended up in the final plans indicates a significant error in the design process and a failure of the review process used by the engineers responsible for the design. The design review process itself, particularly when renovations are made to an existing bridge, has become the focus of the investigation.
Speaking of presentations, I'd love to hear one of our alumni walk through the part of the calculation I know nothing about, the stress calculation and the treatment of rivets in the net force and torque calculation.
The most striking thing to me was the similarity between the way they presented their results and the way the West Point Bridge Design computer program works to highlight strong and weak points in a design, although this particular design element is not part of the simple bridge designs used in that program.
Wednesday, May 2, 2007
How To Do an Engineering Problem
George Heller, shown at left, is an alumnus of TCC and a current student in the Mechanical Engineering department of the FAMU-FSU College of Engineering. He is also a blacksmith.
George is acting as the liason between TCC and the FSU ASME chapter this year, so he attends most club meetings. He will gladly share what he has learned there, as he did last week.
Information presented at the 20 April meeting:
Before the meeting even began, George shared one of the things he learned this semester: How to do a problem, engineer style. He was pleased to point out that some of the steps are ones he learned in my class, but his real point was that a correct answer would not be given full points if it was found while skipping some steps that a physicist like me will normally omit. That may be why he was laughing.
Side comment: I suspect he never took a class from Doug Jones. Students who have taken a class from Doug Jones will understand this comment, although even Doug does not emphasize some of those points because they rarely appear in math problems.
The image above is clickable, but the one below might be more readable. The point he emphasized the most was step 1. There is no explicit penalty for not reading each word (emphasis added by George in the original), but you might get zero for a problem if you miss a key adjective or participle. A careful, close reading of each problem is crucial, and might be a point I will emphasize in the fall.
I already emphasize step 2, primarily as a result of past conversations with engineering faculty about what weaknesses students bring with them from physics classes. George added that you would lose 5 points if the right pictures are not drawn, and drawn well, whether your answer is right or not. Turning words into pictures is as important as turning words into equations (which are steps 3 and 4). He added that you would lose one point for each given value you did not specify (with units) in step 3, and lose a point if your specification of the equations was incomplete in step 4.
Steps 5, 6, and 7 (what students correctly consider "solving" the problem) should be automatic by the time you get out of TCC, so you will get the right answer if you do the first parts correctly. That is, in fact, the reason for the emphasis on the work that has to be done before you can "solve" the problem.
Finally, a point will be taken off if you do not box your answer.
Thanks, George!
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