Radio Communication Between the Stars

Posted on 29-August-2013 by mathscinotes

Introduction

A phone call from a cancer epidemiologist a few months ago got me thinking radio communication over long distance. This researcher was a sharp guy who immediately saw the dynamic range issues associated with cell tower communication -- the issues are even worse for space communication.

I have always had a keen interest in radio communication. Two-way radios, for example, have so many unique advantages as they can be used where mobile phone reception is not reliable and they can also be used to ensure consistent and secure lines of communication between construction site workers and even the military. I know that you can find 2 way radios for sale in NYC, NJ & PA and many other states too, so, if you are looking for a walkie talkie, be sure to do plenty of research first to find the right one for you.

I recalled that I had archived some interesting paper articles years ago in my paper files (of course, indexed by an Access database I also wrote years ago). In these files I found an interesting Scientific American article on the difficulties involved with radio communication between the stars -- the article was written by a SETI researcher. There is "mathematical gold" to be mined in this article. This post just reviews the mathematics in this article. Let's dig in ...

Background

The article addresses three interesting types of problems:

  • Estimating the Number of Stars Near the Sun

    This problem involves making an estimate of the star density in the vicinity the Sun by using the Gliese Catalogue of Nearby Stars. Given the star density, we can compute the number of stars we would expect to find in a sphere of 100 light-years radius about the Sun.

  • Computing the Power Required to Send A Clear Signal to All Stars Near the Sun Simultaneously

    Here we compute the power required to drive a signal powerful enough to be received by all stars within 100 light-years distance of the Sun using an antenna that radiates power equally in all directions (i.e. isotropic radiator).

  • Computing the Power Advantage of Beamforming

    Using a large antenna, we can generate a very directional beam. This allows us to reduce our instantaneous power needs enormously, but we can only broadcast to a very narrow region of the sky. I have gone into beamforming basics in this post.

Analysis

Number of "Local" Stars

The analysis is pretty straightforward:

  • Get a catalog of nearby stars.

    There are numerous catalogs available -- just pick one that focuses on local stars.

  • Find out how many stars are near the Sun.

    I will use a 5 parsec (pc) neighborhood of the Sun to estimate the star density.

  • Estimate the density of stars in the Sun's neighborhood.

    I will assume this density is the same out to 100 light-years from the Sun. The idea here is that (1) we are more likely to have a good accounting of stars near the Sun that those further out because many stars are dim, and (2) the galaxy is so large relative to 100 light-years that the density of stars probably changes little out to 100 light-years.

  • Estimate the number of stars in a 100 light-year neighborhood of the Sun.

    Simply multiply the volume in a 100 light-year sphere by our estimate for the density of stars.

Figure 1 shows my Mathcad worksheet that estimates the number of stars within 100 light-years of the Sun. This analysis follows the approach of this source.

Figure 1: Estimate for the Number of Stars Within 100 Light-Years of the Sun.

Figure 1: Estimate for the Number of Stars Within 100 Light-Years of the Sun.

Definition of Parsec NASA reference calculation

Thus, there are likely over 14,000 stars within 100 light-years of our Sun.

Power for Driving an Isotropic Radiator over 100 Light-Years.

The article author computed the power needed to drive a readable signal based on the following assumptions:

  • Assume the antenna is an isotropic radiator.

    This assumption means that the radio power is projected equally in all directions -- the radio power distribution is spherically symmetric.

  • Assume the receive antenna has aperture of 1 m2.

    We have to make some assumption for the size of the receive antenna in order to estimate the amount of transmit power that will be received.

  • The power is readable at 100 light-years distance.

    Given a 1 m2 antenna aperture, this means that the signal level at 100 light-years is at the thermal noise level.

Figure 2 contains the calculation, which also shows that this transmit power is more than 5000 times greater than the total electrical power generation capacity of the United States.

Figure 2: Calculate the Power Needed to Communicate with a Star 100 Light-Years Away.

Figure 2: Calculate the Power Needed to Communicate with a Star 100 Light-Years Away.

Definition of Parsec Total Power Generation Capacity

Beamforming's Power Advantage

Figure 3 shows the calculations associated with estimating the power advantage gained by using a large antenna with a very narrow, circular beam. This beam would be tough to steer accurately, but it would make the amount of power required substantially less. Unfortunately, you probably would only be able to send a signal to one star at a time.

Figure 3: Power Advantage Provided By Beamforming.

Figure 3: Power Advantage Provided By Beamforming.

Circular Aperture Model

Conclusion

I have reviewed the calculations in the article and I was able to duplicate the author's results. This analysis shows the difficulty associated with radio communication between stars -- it is more than slow -- it takes power and/or an extensive antenna system.

Posted in Astronomy, General Mathematics | 7 Comments

Battleship Rangefinders and Geometry

Posted on 23-August-2013 by mathscinotes

Quote of the Day

Russia is a riddle wrapped in a mystery inside an enigma.

— Winston Churchill.

Introduction

While reading a web page on WW2 naval warfare, I found some interesting material on how naval gunfire was spotted -- the process for correcting aiming errors. This web page contained Table 1, which indicates the maximum range at which an observer 100 feet above the waterline on a ship (called "own ship") can see another ship (called "target") of a given height above the waterline.

Table 1: US Navy Table of Maximum Range Assuming a 100 foot Rangefinder Height and Variable Target Height.

Target Height(ft)

Range (yards)

10

30700

20

33600

30

36100

40

37900

50

39700

60

41300

70

42800

80

44200

90

45400

100

46600

110

47600

120

48900

I can regenerate Table 1 using the distance-to-the-horizon equation I derived in this post. I thought this would be a nice application example.

Let's dig in ...

Background

Table 1 assumes that we are limited to seeing objects that are visible optically on the horizon assuming refraction. Before radar, battleships had all sorts of optical measuring instruments for detecting targets and measuring their range. Figure 1 shows a battleship photograph that calls them out (Source).

Figure 1: Optics Deployed on the USS Idaho.

Figure 1: Optics Deployed on the USS Idaho.

For battleships to engage targets at long range, they needed to be able to accurately determine the range to targets and optical rangefinders were an excellent solution prior to the arrival of radar. In this post, I want to take a closer look at how rangefinders worked. The original concept for these naval optical rangefinders was developed by Bradley Fiske, a naval technologist that few people know about today. I read his biography and was very impressed with the scope of his inventiveness.

There were many variations on the basic rangefinder idea. For my purposes here, we will look more closely at the variant known as a coincidence rangefinder. Figure 2 shows a block diagram of a coincidence rangefinder (Source).

Figure 2: Block Diagram of a Coincidence Rangefinder.

Figure 2: Block Diagram of a Coincidence Rangefinder.

There are two prisms (actually called pentaprisms) at the end of two long arms -- each arm often 4 or 5 meters in length. The design brings the images into coincidence by rotating the compensating wedge prisms. In theory, they could have rotated one of the pentaprisms to bring the images into coincidence, but it turns out that is difficult to do accurately on targets at long range. The compensating wedge prisms provide a more accurate solution. There are numerous subtleties in the construction of these devices that are beyond the scope of this note. See the source for some of the interesting details. The compensator prism wedges and their function are particularly interesting.

These rangefinders were quite large and were often armored. Figure 3 shows a rangefinder from the Graf Spee, a German "pocket" battleship (Source).

Figure 3: Rangefinder from the Graf Spee.

Figure 3: Rangefinder from the Graf Spee.

The following quote nicely describes its function (Source).

The coincidence range finder uses a single eyepiece and uses a prism to merge images from both lenses into a single image to present to the operator. The operator adjusts the rotation of the prisms using a dial until the images overlap in the eyepiece. The degree of rotation of the prisms determines the range to the target by simple trigonometry.

The operation of the pentaprisms is interesting. Figure 4 contains a nice illustration of how the light bounces around a pentaprism (Source).

Figure 4: Light Bouncing in a Pent-Prism.

Figure 4: Light Bouncing in a Pentaprism.

The basic geometry involved with a coincidence rangefinder is simple enough. Figure 5 shows that the coincidence rangefinder determines range using simple Euclidean trigonometry (Source).

Figure 5: Coincidence Rangefinder Geometry.

Figure 5: Coincidence Rangefinder Geometry.

Here is another drawing that does a nice job of illustrating how the rangefinders were constructed (Source).

Figure 6: Top View of an Optical Rangefinder.

Figure 6: Top View of an Optical Rangefinder.

For those of you interested in how the rotating prism can support precise angular measurements, see this comment for details.

Analysis

Strategy

Here is my analysis approach:

  • Develop a formula for the maximum range of an optical rangefinder

    I will assume that the observation height above the waterline is what limits the maximum range of the rangefinder.

  • Determine the refraction radius that the US Navy used.

    The maximum range will be affected by refraction. This post gives us a way to model the refraction by using a refraction radius.

  • Compare my results with the results listed by the US Navy.

    I will examine the differences between the US Navy's table and the equivalent table generated by my formula.

Optical Rangefinder Maximum Range Formula

Equation 1 shows the formula for determining the range to the horizon assuming refraction (from this post).

Eq. 1 {{s}}\left( h,{{r}_{E}},{{r}_{R}} \right)=\sqrt{\frac{2\cdot {{r}_{E}}\cdot h}{1-\frac{{{r}_{E}}}{{{r}_{R}}}}}

where

  • rE is the radius of the Earth.
  • rR is the radius of the refraction circle.
  • h is the height of the observation point.
  • s is the arc length from the observation point to the horizon.

Figure 7 shows the maximum range scenario that a battleship would experience with a target. We will apply Equation 1 twice to determine the maximum rangefinder range.

Figure 7: Optical Geometry for Analysis.

Figure 7: Optical Geometry for Analysis.

Using Figure 6 and assuming each ship just sees the other on the horizon, we can determine the maximum range using Equation 2 -- which uses Equation 1 twice.

\displaystyle {{s}_{Rangefinder}}\left( h,H,{{r}_{E}},{{r}_{R}} \right)=\sqrt{\frac{2\cdot {{r}_{E}}\cdot h}{1-\frac{{{r}_{E}}}{{{r}_{R}}}}}+\sqrt{\frac{2\cdot {{r}_{E}}\cdot H}{1-\frac{{{r}_{E}}}{{{r}_{R}}}}}

Eq. 2 \displaystyle {{s}_{Rangefinder}}\left( h,H,{{r}_{E}},{{r}_{R}} \right)={{s}_{O}}+{{s}_{T}}
\displaystyle {{s}_{Rangefinder}}\left( h,H,{{r}_{E}},{{r}_{R}} \right)={{s}}\left( h,{{r}_{E}},{{r}_{R}} \right)+{{s}}\left( H,{{r}_{E}},{{r}_{R}} \right)

where

  • H is the height of the rangefinder.
  • h is the height of the target.
  • sO is the arc length from my own ship to the horizon.
  • sT is the arc length from the target ship to the horizon.

US Navy Refraction Radius

This is an error minimization problem. I can write a routine that finds the refraction radius that minimizes the error between my formula (Eq. 2) and the US Navy's table. Figure 8 shows my Mathcad routine form performing this minimization. I minimized the maximum error. You could also minimize the maximum percentage error -- the answers are very similar.

Figure 8: Determination of Refraction Radius Used By US Navy.

Figure 8: Determination of Refraction Radius Used By US Navy.

Error Between My Formula and the US Navy Table

Figure 9 shows my comparison between the US Navy result (Table 1) and my formula (Eq. 2). The agreement is very good considering they probably used slide rules for their work. I shudder thinking of my own youth when I had to use slide rules.

Figure 9: Comparsion by US Navy Rangefinder Table and My Formula.

Figure 9: Comparison by US Navy Rangefinder Table and My Formula.

Conclusion

I showed that the US Navy table and my formula produce results that are within 0.3% of each other for a refraction radius that is 6.975 that of the Earth's radius (they probably just used 7). This accuracy is reasonable assuming they were using a slide rule for their computations.

In Figure 10, I thought I would add an excellent picture of a rangefinder from a US Navy ship's fire control system to show how their appearance could vary (Source).

Figure 10: Rangefinder incorporated in Mk19 Fire Control System on USS Pennsylvania (BB 38).

Figure 10: Rangefinder incorporated in Mk19 Fire Control System on USS Pennsylvania (BB 38).

If you are looking for more information on WW2 rangefinders, here is a manual. It is a large file (>7MB).

Save

Save

Save

Save

Save

Save

Save

Posted in Naval History, optics | 19 Comments

Distance to the Horizon Assuming Refraction

Posted on 19-August-2013 by mathscinotes

Quote of the Day

Fate rarely calls upon us at a moment of our choosing.

— Optimus Prime. This quote shows that inspiration can come from anywhere.

Introduction

Figure 1: Illustration Showing that Refraction Increases the Perceived Distance to the Horizon.

Figure 1: Illustration Showing that Refraction Increases the Perceived Distance to the Horizon.

While doing some reading on lighthouses, I needed a formula to compute the distance to the horizon as a function of height. This formula would give me an idea of how far away a lighthouse could be seen.

As I usually do, I started with the Wikipedia and it listed two interesting equations:

  • Eq. 1: \displaystyle s\left[ \text{km} \right]=3.57\cdot \sqrt{h\left[ m \right]}

    where s is the distance along the Earth's surface in kilometers and h is the height of the observer in meters. This formula assumes that light moves strictly in straight lines. You will also see it in the form s=\sqrt{2 \cdot h \cdot r_E}, where r_E is the radius of the Earth.

  • Eq. 2: \displaystyle s\left[ \text{km} \right]=3.86\cdot \sqrt{h\left[ m \right]}

    This formula assumes that light is refracted and travels along a circular arc.

Allowing for refraction means that light will curve around the geometrical horizon (Eq. 2),  which means that we will see objects that are just beyond the horizon. Using Eq. 2, we will compute a horizon distance that is about 7% further than predicted by Eq. 1. Figure 1 illustrates this point.

While I frequently see refraction modeled using Eq. 2, I have not seen anyone go into the details of Eq. 2 -- I am sure this analysis exists -- I just have not seen it. So I will dive into the details of Eq. 2 in this post. I have seen the Eq.1 derived in a number of places, so I will not repeat its derivation here.

The material presented in this blog is similar to that presented on mirages. I have included some mirage information in Appendix B.

Background

Basic Idea

Light bends because of density differences in the atmosphere. These density differences are caused by (1) altitude and (2) temperature. Of course, temperature and pressure vary with time. To get some idea of the effect of refraction on the distance to the horizon, we need to assume a "typical" atmosphere. We can derive Eq. 2 using the typical atmosphere assumption, but we need to understand that those conditions are often not present.

Analysis

Here is my problem solving approach:

  • Show that the air's index of refraction is a simple function of density.
  • Develop an expression for the rate of change of air's index of refraction.
  • Develop an expression for the radius of curvature of a refracted light beam.
  • Develop an expression for the arc length along the Earth's surface that a refracted light beam will traverse.
  • Simplify the expression by assuming typical atmospheric conditions and common units.

Air's Refractivity as a function of Density

Because air's index of refraction is very close to 1, physicists usually talk in terms of refractivity. The refractivity of air is defined as \displaystyle {{N}_{Air}}\triangleq \left( n-1 \right)\cdot {{10}^{6}}, where n is the index of refraction. One common formula for the refractivity of dry air is given by Equation 3 (Source).

Eq. 3 \displaystyle {{N}_{Air}}=\left( 776.2+\frac{4.36\cdot {{10}^{-8}}}{{{\lambda }^{2}}} \right)\cdot \frac{P}{T}

where

  • P is air pressure [kPa].
  • T is the air temperature [K].
  • \lambda is the wavelength of light [cm].

Using the ideal gas law, we can show that temperature and pressure are closely related to density as shown in Equation 4.

Eq. 4 \displaystyle P\cdot V=n_m\cdot R\cdot T\Rightarrow \frac{P}{T}=\frac{m}{M{{W}_{Air}}}\cdot \frac{R}{V}=\frac{m}{V}\cdot \frac{R}{M{{W}_{Air}}}=\rho \cdot \frac{R}{M{{W}_{Air}}}

where

Equation 4 states that the ratio of P/T is proportional to \rho (R and MWAir are constants). This means we can restate Equation 3 in the simpler form shown in Equation 5.

Eq. 5 \displaystyle {{N}_{Air}}=a\left( \lambda \right)\cdot \rho

where

  • a\left( \lambda \right) is a function of the wavelength of light.

The index of refraction is a function of wavelength, as is shown in Figure 2. Note how the shorter wavelengths have a larger index of refraction than the longer wavelengths.

Figure 2: Air's Index of Refraction Versus Wavelength.

Figure 2: Air's Index of Refraction Versus Wavelength.

For this post, I will assume that the wavelength of light is fixed and that a\left( \lambda \right) is a constant.

Air's Index of Refraction Rate of Change

For light to refract in the atmosphere, it must encounter air with an index of refraction that varies. The air's index of refraction is a function of its density, which varies with altitude and temperature. We can express the variation of the atmosphere's index of refraction using Equation 6.

Eq. 6 \displaystyle \frac{dn}{dz}=-\frac{\left( n-1 \right)}{T}\cdot \left( \frac{\rho \cdot g\cdot T}{P}-\gamma \right)

where

  • n is the air index of refraction.
  • z is the altitude.
  • g is the acceleration due to gravity.
  • \gamma is the atmospheric lapse rate, which is defined as the rate of temperature decrease with height (i.e. \gamma =-\frac{dT}{dz}).

Deriving this expression requires manipulation of the ideal gas law. The derivation is straightforward but tedious. I have included it in Appendix A.

Equation 6 has a couple of interesting aspects:

      • \frac{dn}{dz}=0 when \gamma = \frac{\rho \cdot g\cdot T}{P}

        This value of \gamma is known as the autoconvective lapse rate. This rare condition means that light does not refract. It occurs over surfaces that are easily heated, like open fields and deserts. This condition requires that the rate of temperature decrease with altitude to cause an air density increase that exactly cancels the air density decrease with altitude.

      • Usually, \frac{dn}{dz} < 0

        The change of index with altitude is normally negative because the air density increase with altitude associated with decreasing temperature is not enough to cancel the air density decrease with altitude. Since light bends towards regions of higher density, light projected straight from a lighthouse will refract toward the Earth's surface under normal circumstances (as shown in Figure 1).

We can compute the autoconvective lapse rate as shown in Figure 3. The value I obtain here can be confirmed at this web page.

Figure 3: Calculation of the Autoconvective Lapse Rate.

Figure 3: Calculation of the Autoconvective Lapse Rate.

Standard Atmosphere

Radius of Curvature for the Refracted Light Beam

Assuming a typical atmosphere, we can model the path of a refracted beam of light in the atmosphere as an arc on a circle. Figure 4 shows a derivation for the radius of curvature for a refracted beam of light. The radius of curvature will be constant (i.e. a circle) when \frac{dn}{dz} is a constant.

Figure 4: Derivation of the Radius of Curvature Expression.

Figure 4: Derivation of the Radius of Curvature Expression.

In the case of a lighthouse beam, \theta =90{}^\circ. This means we can write the refraction radius as \displaystyle {{r}_{R}}=-\frac{n}{\frac{dn}{dz}}.

Arc Length Traversed By A Refracted Light Beam

We are going to model the refracted light beam as moving along the arc of a circle of radius larger than that of the Earth. Figure 5 illustrates this situation.

Figure 5: Illustration of the Refracted Light Beam Moving Along the Arc of a Circle.

Figure 5: Illustration of the Refracted Light Beam Moving Along the Arc of a Circle.

Given the geometry shown in Figure 4, we can derive an expression for the arc length upon the Earth's surface that the light beam traverses as shown in Figure 6.

Figure 6: Derivation of Formula for the Distance Traveled Along the Earth's Surface By Refracted Light Beam.

Figure 6: Derivation of Formula for the Distance Traveled Along the Earth's Surface By Refracted Light Beam.

Equation 7 shows the key result from the derivation in Figure 6.

Eq. 7 \displaystyle {{s}_{R}}\left( h,{{r}_{E}},{{r}_{R}} \right)=\sqrt{\frac{2\cdot {{r}_{E}}\cdot h}{1-\frac{{{r}_{E}}}{{{r}_{R}}}}}

where

  • rE is the radius of the Earth.
  • rR is the radius of the refraction circle.
  • h is the height of the lighthouse.

Notice how Eq.7 can be viewed as Eq. 1 with an enlarged Earth radius of {{{r}'}_{E}}=\frac{{{r}_{E}}}{1-\frac{{{r}_{E}}}{{{r}_{R}}}}. I see refraction often modeled by engineers who simply use Eq. 1 with an enlarged Earth radius. The radius used depends on the wavelength of the photons under consideration.

Simplified Arc Length Expression

Figure 7 shows how I obtained Equation 2 from Equation 7.

Figure 7: Equation 2 from Equation 7.

Figure 7: Equation 2 from Equation 7.

_AirIndex

Conclusion

While the derivation was a bit long, I was able to derive Equation 2 from first principles. The derivation shows that the final result is sensitive to the choice of lapse rate, which varies throughout the day. I should note that the use of a refraction radius that is a multiple of the Earth's radius is often applied to other types of electromagnetic signals. For example, radar systems frequently use a "4/3 Earth radius" for refraction problems (Source). The refraction radius for radar is different than for optical signals because the index of refraction in the radio band is different from that of optical signals.

Appendix A: Derivation of Rate of Change of Atmospheric Index of Refraction with Lapse Rate.

Figure A: Derivation of Rate of Change of Atmospheric Index of Refraction with Lapse Rate.

Figure A: Derivation of Rate of Change of Atmospheric Index of Refraction with Lapse Rate.

Ideal Gas Law Density of an Ideal Gas Law Definition of Lapse Rate Ref on Atmospheric Optics

Appendix B: Excellent Discussion of Index of Refraction, Lapse Rate, and Mirages.

I really like the way this author discusses mirages (Source).

Figure B: Excellent Description on the Formation of Mirages.

Figure B: Excellent Description on the Formation of Mirages.

Save

Save

Save

Save

Posted in General Science, Navigation, optics | 26 Comments

The Drinking History of the US

Posted on 12-July-2013 by mathscinotes

I have been reading "A Wicked War: Polk, Clay, Lincoln and the 1846 U.S. Invasion of Mexico" and I notice a common theme with other history books about the early US --drinking alcohol was a major preoccupation with early Americans. I became curious enough that I looked in the book "The Alcoholic Republic: An American Tradition" by W.J. Rorabaugh and found the following chart.

All_Alcohol

Notice the huge drop in consumption around 1850. This drop is consistent with development of a temperance movement in the US in the mid-19th century. This movement was significant enough to be mentioned in the history books that I have been reading.

There is also the drop around the 1920s to 1930s, that is when prohibition was put into effect. Notice how it doesn't completely drop to 0. That's because prohibition did not (and could not) stop people from consuming alcohol. Blood alcohol tests like those utilized when drug testing in Pampa TX would not be in common use by law enforcement until after prohibition. Meaning even if prohibition outlawed the drinking of alcohol, it would have been very hard to prove someone had done so unless they were visibly drunk. Drinking levels were low, but it was never 0.

As a boy, I delivered the Minneapolis Star afternoon newspaper door-to-door. Every two weeks I would collect money for the papers I delivered. People usually invited me into their homes because Minnesota was often cold outside. I was always surprised at the number of people who were drunk when I entered their homes. It did not matter what time of day I stopped by. A few of them were my peers so must have known where to buy fake id cards so they could buy alcohol underaged. I cannot imagine what I would have encountered if the alcohol consumption levels were two or three times greater than they were in 1970, a year in which I delivered papers. According to the 2018 National Survey on Drug Use and Health, 14.4 million adults (18 and above) had Alcohol Use Disorder. I guess we have to grateful to those who run and work in addiction treatment centers, like Enterhealth (see their center at https://enterhealth.com/), as they help these people overcome their struggles with alcohol and addiction.

Posted in History of Science and Technology, Osseo, Personal | 3 Comments

Torpedo on an Aruba Beach

Posted on 12-July-2013 by mathscinotes

I have read in a number of places about torpedoes being found on beaches during wartime, but I have never seen pictures of one on a beach. While researching possible diving locations, I ran across this excellent excellent web article on a German WW2 torpedo that ran up on a beach in Aruba. Here is a link to a copy of contemporaneous news article on the same subject.

I have included the best picture I could find below.
Aruba
I would not stand that close to any unexploded ordnance. The text in the photograph indicates four men were killed trying to disarm it.

I have only had one hazardous incident in my career and it involved an underwater vehicle. During testing of a fully fueled vehicle, a thermal battery failed spectacularly immediately after I pushed the start button. The umbilical cable to the vehicle was blown free and a tongue of flame emerged from the vehicle's umbilical connector. This was not good. As always, I immediately suspected the battery had failed, which was mounted at the very end of the vehicle forebody (i.e. front-half of the vehicle). We knew the battery used the fuel tank in the afterbody as a heat sink, so another engineer and I quickly separated the front and rear sections of the vehicle. This action separated the battery from the fuel tank. Looking inside the vehicle forebody, I could see that the battery was so hot it was glowing red -- I could feel the heat on my face. Luckily, the fuel tank in the afterbody, filled with Otto fuel, had not been damaged. Otherwise, the fuel tank could have exploded and the coroner would have been asking for my dental records.

We contacted the battery vendor and they performed an analysis on the failed battery. Apparently, the battery had the plus and minus terminals swapped. Of the 15 batteries we had on hand, only one had this problem. This was scary. The battery had so much energy in it that it blew a large hole in our power supply circuit board.

I have always disliked batteries and "death by battery" is not how I want to go. I shudder just thinking about it ...

Posted in Naval History | Comments Off on Torpedo on an Aruba Beach

Fiber Optic Deployment Woes

Posted on 11-July-2013 by mathscinotes

We recently had a customer who reported that one of our products was reporting low RF video output power. This normally is caused by an issue with equipment setup, but everything this customer did was fine. It turned out that this particular unit had a serious ant problem in a fiber splice tray, a small box used to protect splices done on multiple optical fibers. After we cleaned out the ants, everything worked fine. How the ants caused a fiber failure is something we are still looking into. The figure below shows that the ants had found a nice, warm home.

Ants

Posted in Fiber Optics | Comments Off on Fiber Optic Deployment Woes

Geothermal Power Math

Posted on 7-July-2013 by mathscinotes

Quote of the Day

Books are standing counselors and preachers, always at hand, and always disinterested; having this advantage over oral instructors, that they are ready to repeat their lesson as often as we please.

- Louis Nizer

Introduction

Figure 1: Strokkur Geyser in Iceland.

Figure 1: Strokkur Geyser in Iceland. Geysers are powered by the Earth's internal heat. The Earth remains hot because of radioactivity and the residual heat of formation. (Source)

Everyday I visit the site refdesk.com because it contains little facts and figures that someone like me really gets into. A few days ago, the following piece of trivia was in their "Fact of the Day" section [Source-item #5]:

Holes drilled as deep as 5 miles into the Earth's surface reveal that the rock temperature increases about 37 °F per 320 feet.

You usually see this number stated as 20 °C/km (example). This piece of trivia caught my eye because it is the Earth's temperature gradient near the surface. If it is relatively constant around the world, we should be able to use this number to compute the total heat output of the Earth. I did a bit of research, and I quickly discovered that there are many nuances to working with geophysics. I also learned that geophysics is a really interesting field with a lot of great mathematics. Unfortunately, it is a field where direct measurements of significant Earth features, like the core, cannot be obtained. It may be the case that Earth is a billion years old, but very little is actually known about such an important bit of it as the core. It is difficult to obtain even basic properties of materials (like the density of iron) at the temperatures and pressures that exist at the Earth's core.

Even with these difficulties, I thought I could do a bit of math to come up with rough approximations for some of the numbers I encountered in my research. Let's look at two numbers:

  • PTotal = 44 to 47 TerraWatts (TW): The total heat output of the Earth.

    The Earth has multiple sources for its internal heat: radioactive decay (see this blog post), the initial heat present since the formation of the Earth, latent heat from the solidification of the core, tidal heating, etc. If we know that temperature gradient near the Earth's surface, we should be able to estimate the total amount of heat being generated from these sources within the Earth.

  • Solidification rate = 1 mm/year: The rate at which the inner core is solidifying.

    The Earth is slowly cooling. One result of this cooling is that the solid inner core is slowly growing.

One concept that intrigues me is the idea that the Earth would be warm underground even without the Sun present. See the Wikipedia for an interesting discussion of this topic. My favorite science fiction story is After Worlds Collide, which is tale that includes a rogue planet called Bronson Beta. This rogue planet survived a very long trip through the bitter cold of interstellar space. Its former inhabitants had built deep underground tunnels that provided a warm sanctuary for travelers from Earth.

Background

Structure of the Earth

Figure 2 shows the layered structured of the Earth's interior (Source).

Figure 1: Layered Structure of the Earth.

Figure 2: Layered Structure of the Earth.

We can only take direct measurements of the rocks down to about 12 km. In fact, we have never drilled deep enough to reach the mantle, though we are trying. We can measure things like temperature gradients near the surface and we can also analyze the rumblings from within the Earth to determine the basic structure of the layers.

Analysis

Power Flow and Temperature Gradient

Equation 1 relates power flow to the temperature gradient and the crust's thermal conductivity.

Eq. 1 \displaystyle q=\lambda \cdot \frac{\delta T}{\delta d}

where

  • q is the heat flow density [mW/m2]
  • ? is the thermal conductivity of the material being measured [W/(m K)]
  • \displaystyle \frac{\delta T}{\delta d} is the temperature gradient [K/m] with respect to depth (d)

There are a number of difficulties in applying Equation 1 over the Earth's surface. A major one is the wide variation in the thermal conductivity of the crust. Generally, we work with averages. Figure 3 shows a graph of the thermal conductivity of crustal rocks versus temperature (Source).

Figure 2: Thermal Conductivity of the Crust as a Function of Temperature.

Figure 3: Thermal Conductivity of the Crust as a Function of Temperature.

The Wikipedia uses a value of 3.0 W/(m K) for the thermal conductivity of the continental granite. They use this value to estimate the heat flow from a square meter of continent as shown in Equation 2.

Eq. 2 \displaystyle {{q}_{Continent}}\approx 3.0\left[ \frac{\text{W}}{\text{m}\cdot \text{K}} \right]\cdot 20\left[ \frac{20\text{K}}{1000\text{ m}} \right]=60\frac{\text{mW}}{{{\text{m}}^{2}}}

This is a rough calculation. Much more precise measurements and calculations have been performed for the continents and the ocean. Table 1 contains a summary of the power density results and the total geothermal power estimates of different researchers. My work here will use the results from Pollack et al (highlighted) because that is what I find most frequently used in popular discussions.

Table 1: Summary of Geothermal Power Density and Total Power Estimates by Researcher and Year of Publication.
Study Year Continental (mW-m-2) Oceanic (mW-m-2) Total (TW)
Williams and von Herzen 1974 61 93 43
Davies 1980 55 95 41
Sclater et al. 1980 57 99 42
Pollack et al. 1993 65 101 44
Labrosse 2007 65 94 46

Total Power Output

Given the average power density from the continents and the ocean, we can estimate the total power generated within the Earth. This does require that we determine the percentages of continental crust and oceanic crust. For my estimates, I assumed that 40% of the Earth's surface is continental crust (Source). Figure 4 shows my calculations, which includes adding 3 TW for the power from mantle plumes. I saw that some authors made this assumption to model places on the Earth's surface like Yellowstone, which have much higher than average heat flux.

Figure 3: Calculation of the Total Heat Production from The Earth.

Figure 4: Calculation of the Total Heat Production from The Earth.

Density of Iron at the Core Good physics page on the Earth Iron Heat of Fusion

My result of 47 TW is a bit higher than those reported in Table 1, but close enough for my rough work here.

Core Solidification Rate

I saw the 1 mm/year solidification rate quoted from a number of sources. Here is a quote from one of these sources.

Over billions of years, Earth has cooled from the inside out causing the molten iron core to partly freeze and solidify. The inner core has subsequently been growing at the rate of around 1 mm year as iron crystals freeze and form a solid mass.

Note that I also saw estimates like 0.3 mm/year and 1 cm/year stated, but rarely so that I will ignore them here. Figure 5 shows my estimate for the latent heat released when 1 mm/year of iron at the inner core boundary solidifies.

Figure 4: Calculation of the Core Heat Generated for a 1 mm Solidification Rate.

Figure 5: Calculation of the Core Heat Generated for a 1 mm Solidification Rate.

Density of Iron at the Core Good physics page on the Earth Iron Heat of Fusion t TW Usage Example

My 4 TW number agrees with this source. There is some debate about this number and you will see different values used by different authors.

Conclusion

This was an interesting exercise for me. I have always wondered the amount of geothermal energy available on the Earth. The total amount of solar power at the Earth' surface is compared to the total amount of geothermal energy available in Figure 6. There is much more solar power available than geothermal.

Figure 5: Comparison of Total Solar Versus Geothermal Energy Amounts.

Figure 6: Comparison of Total Solar Versus Geothermal Energy Amounts.

Solar insolation constant Google search for Earth's radius Wikipedia page on Earth's energy budget

Save

Save

Save

Save

Save

Posted in Geology, History of Science and Technology | 1 Comment

GPAs and Work Performance

Posted on 5-July-2013 by mathscinotes

I have spent a lot of time interviewing engineers. In my current job, the first employee in the hardware department was me and I have hired every hardware person at this site. I have spent a lot time thinking about how to pick the right people (e.g. blog post, blog post). The New York Times recently published an article that contains an interview with a Google executive and he made the following statement.

One of the things we've seen from all our data crunching is that G.P.A.'s are worthless as a criteria for hiring, and test scores are worthless - no correlation at all except for brand-new college grads, where there's a slight correlation. Google famously used to ask everyone for a transcript and G.P.A.'s and test scores, but we don't anymore, unless you're just a few years out of school. We found that they don't predict anything.

I completely agree with this statement. After decades of interviewing people, I believe there is minimal correlation between grades in school and performance on the job. A couple of examples will help illustrate my point. One of the finest engineers I have ever worked with barely passed engineering school. I met him at HP (back in the days of Bill and Dave). Normally, you could not get an interview at HP without at least a 3.5 GPA. He was an absolute wizard -- everyone went to him for circuit advice and he always helped. As an electrical engineer, he understood circuits really well and he would often fix them and make sure they were working properly. He knew about a lot of Interesting technology including RAFI Control Components and he would regularly use them to create controlled circuits. He worked so hard and such long hours that he even had a hammock in his cube. What I admired most about him was his kindness and willingness to help others. He made all around him better. I work hard to emulate his way with people in need of help. Don't worry, he was rewarded for his hard work, and we always made sure that the company survey questions on the employee feedback forms gave plenty of opportunities to report any dissatisfaction that anyone had with their experiences in the workplace. In a way, he was quite efficient in handling the work pressure that came with the task he was assigned. It is relieving to learn that a lot of corporate offices make use of the workload management tool to avoid employee burnout.

On the other hand, while managing a software group at another company, I worked very hard to fire a PhD who never had less than an "A" in his life. One of my staff used to refer to him as "heroically lazy". Never in my career had I seen a human being work so hard at not working. He was not fun to speak with -- every conversation with him had a tone of condescension and entitlement. I have many stories of his epic quest to avoid any form of work. Thankfully, this issue was brought to my attention and I was able to fire him before he wasted any more of our resources. To make sure this sort of thing isn't happening at your business, it might be worth using the software at https://www.qualtrics.com/employee-experience/360-degree-feedback/ to gather reviews about each employee from peers and other managers. This will allow people to explain any issues that this person may be causing, allowing you to remove them from the workplace.

The cases cited here are not typical and I do not mean to imply that there is an inverse correlation between grades in school and work performance. I am saying that school performance and work performance are uncorrelated. I do believe that grades identify people who perform well in the academic environment, thus they help academia identify its best and brightest. However, academia and a commercial engineering firm are two very different worlds.

I have not come up with an unambiguous formula for identifying engineering talent -- such a formula does not exist. However, I know that GPA alone is not the answer.

Posted in Management | Comments Off on GPAs and Work Performance

Granite Self-Heating Math

Posted on 3-July-2013 by mathscinotes

Introduction

I came across the following statement in an article about the self-heating of the granite in an article about how radioactivity heats the interior of the Earth.

Radioactivity is present not only in the mantle, but in the rocks of Earth's crust. For example, Marone explains, a 1-kilogram block of granite on the surface emanates a tiny but measurable amount of heat (about as much as a .000000001 Watt [1.0E-9 W or 1 nW] light bulb) through radioactive decay.

I will do a quick calculation to verify that I understand where this number originates.

Background

I will assume that all the thermal power generated in a 1 kg granite piece comes from the radioactive decay of uranium and thorium that make up a small part of many granite deposits. It turns out that I have had to deal with the fact that many substances are slightly radioactive throughout my career. For example, memory chips have long been recognized as vulnerable to soft errors from alpha particles originating from trace radioactive materials in electronic packaging. I have spent much time implementing error correcting codes on memory systems in order to deal with single bit upsets.

Analysis

These calculations are similar in principle to those done in this blog post. My approach is simple:

  • Gather information on uranium and thorium concentrations in granite (see Appendix C)
  • Gather information on how uranium and thorium decay (see Appendices A and B, respectively)
  • Perform routine radioactive decay calculations, assuming that all the alpha particle energy eventually ends up as heat.

Setup

Figure 1 illustrates the setup I went through for the calculation.

Figure 1: Setup for the Analysis of Granite's Self-Heating.

Figure 1: Setup for the Analysis of Granite's Self-Heating.

Wikipedia Entry for Uranium

Calculations

Figure 2 shows the actual calculations.

Figure 2: Granite Self-Heating Calculations.

Figure 2: Granite Self-Heating Calculations.

Wikipedia Entry for Natural Radiation from Granite Wikipedia Entry for Thorium

Conclusion

I encountered a statement in an article on the web that says that ~1 nW of power is generated in a 1 kg block of granite. I compute that the value is 0.52 nW, which is about 1 nW. So I have confirmed the number.

Another byproduct of radioactive decay in granite is radon. However, that is a topic for another blog post.

Appendix A: Uranium Isotope Characteristics

Figure 3 is a table from the Wikipedia entry for uranium.

Figure 3: Uranium Isotopes (Wikipedia).

Figure 3: Uranium Isotopes (Wikipedia).

Appendix B: Thorium Isotope Characteristics

Figure 4 is a table from the Wikipedia entry for thorium.

Figure 4: Thorium Isotopes (Wikipedia).

Figure 4: Thorium Isotopes (Wikipedia).

Appendix C: Uranium and Thorium Concentrations in Granite

I obtained my uranium and thorium concentrations in granite from the Wikipedia. Here is the quote with the pertinent text hightlighted.

Some granites contain around 10 to 20 parts per million of uranium. By contrast, more mafic rocks such as tonalite, gabbro or diorite have 1 to 5 PPM uranium, and limestones and sedimentary rocks usually have equally low amounts. Many large granite plutons are the sources for palaeochannel-hosted or roll front uranium ore deposits, where the uranium washes into the sediments from the granite uplands and associated, often highly radioactive, pegmatites. Granite could be considered a potential natural radiological hazard as, for instance, villages located over granite may be susceptible to higher doses of radiation than other communities.[11] Cellars and basements sunk into soils over granite can become a trap for radon gas, which is formed by the decay of uranium.[12] Radon gas poses significant health concerns, and is the number two cause of lung cancer in the US behind smoking.[13]

Thorium occurs in all granites as well.[14] Conway granite has been noted for its relatively high thorium concentration of 56 (±6) PPM.[15]

Posted in Geology | 2 Comments

Voyager 1 and Gliese 445

Posted on 30-June-2013 by mathscinotes

Introduction

I was reading an article about the Voyager 1 space craft nearing the edge of interstellar space. This article was so interesting that I ended up reading a number of articles on the subject (one example) and they all had the same numbers in them:

  • Voyager 1 is leaving our solar system at 3.6 Astronomical Units (AU) per year.
  • Voyager 1 is moving toward the star Gliese 445 (aka AC+79 3888) and will be at the point of closest approach in about 40,000 years
  • The closest Gliese 445 and Voyager 1 will get is about 1.7 light-years

My interest in these numbers was raised when I noticed that Voyager 1's velocity is too low for it to come near Gliese 445 in 40,000 years. It turned out that Gliese 445 is coming to us faster than Voyager is going to it. I was a bit surprised.

Let's see if we can gain some insight into these numbers. I do not want to get into the details of astrometry, so my work will be approximate (i.e. Fermi problem analysis).

Background

The following background material is useful to review before we dive into the numbers.

Voyager 1

Voyager 1 was a space probe launched in 1977 and it is still performing scientific work. It is currently heading out the solar system. Figure 1 shows its path (labeled V1) relative to the orbital plane (Source).

Figure 1: Path of Voyager 1 With Respect to the Orbital Plane.

Figure 1: Path of Voyager 1 With Respect to the Orbital Plane.


After a very long time, Voyager 1 will pass relatively close to Gliese 445. This post will estimate that time and distance.

Gliese Star Catalog

If you are curious how stars get names like this, check out the Wikipedia entry on the Gliese Star Catalog. This star catalogue lists the stars located within 25 parsecs (81.54 light-year) of the Earth.

Star Distances Versus Time

Figure 2 shows a graph of the distances of various nearby stars to Earth as a function of time (Source).

Figure 2: Graph of the Distances of Nearby Stars to the Earth Over Time.

Figure 2: Graph of the Distances of Nearby Stars to the Earth Over Time.


Observe how much the stars move over long time periods. Note also how Gliese 445 is moving towards the Earth, which means it is also coming closer to Voyager 1. The shape of the distance curve is described by a hyperbola. We derive this curve shape as shown in Figure 3.
Figure 3: Derivation of Hyperbolic Form of Gliese's Distance Equation.

Figure 3: Derivation of Hyperbolic Form of Distance Equation.

Analysis

We are going to compute a few interesting numbers:

  • What is the velocity of Voyager 1 in AU per year?
  • How long does Voyager 1 take to travel one light year?
  • How fast is Gliese 445 approaching the Earth?
  • When will Voyager 1 have its closest approach to Gliese 445?
  • How close will Voyager 1 get to Gliese 445?

Calculations

Voyage 1 Velocity in AU Per Year and the Time for It to Travel One Light Year

Figure 4 shows my calculations for the velocity of Voyager 1 in AU per year (3.595 AU/year) and how long it takes for Voyager 1 to travel one light-year (17,600 years). These numbers agree with those in this NASA article.

Figure 4: Voyager 1 Velocity in AU Per Year and Time to Travel One Light-Year.

Figure 4: Voyager 1 Velocity in AU Per Year and Time to Travel One Light-Year.

Voyager 1 interstellar Voyager 1 current status Voyager 1 current status

Gliese 445 Approach Speed to Earth and Time of Closest Approach

I digitized the distance curve for Gliese 445 in Figure 2 and took the derivative of position versus time. Since the graph for Gliese 445 does not show data for today, I will compute the speed for 20,000 years from now. The distance versus time curve looks fairly linear in this region and the velocity should be similar to what we see today (Figure 5).

Figure 5: Calculations for the Velocity of Gliese 445 Relative to the Sun and The Time of Closest Approach.

Figure 5: Calculations for the Velocity of Gliese 445 Relative to the Sun and The Time of Closest Approach.


I compute an approach velocity of 106 km/sec (much higher than Voyager 1's velocity of 17 km/sec). The Wikipedia lists this value as 119 km/sec -- close enough. I compute that the time of closest approach is 46,000 years from now, which is close to the Wikipedia statement of "about 40,000 years". I compute the distance at closest approach as 3.485 light-years, which is close to the Wikipedia value of 3.45 light-years.

Closest Distance of Voyager 1 to Gliese 445

To get an accurate estimate of this distance, I would need to know details on the exact relationship between the courses of Voyager 1 and Gliese 445, which I do not have access to. To get a rough estimate of how close they could get, let's assume that Voyager 1 is moving toward the point of closest approach between Gliese 445 and the Sun (Figure 6). This will give us an answer that is smaller than the true value, but it will be close enough for my purposes.

Figure 6: Calculation of Closest Approach of Voyager 1 to Gliese 445.

Figure 6: Calculation of Closest Approach of Voyager 1 to Gliese 445.


My result shows that if Voyager was perfectly aimed at Gliese 445, it would get within 0.9 light-years of the star. NASA reports that the distance at closest approach is 1.7 light-years, a number for which I assume they took into account the actual aiming error. My simple result is not too far off.

Conclusion

Just a quick exercise to make sure that I understand what I am reading. The idea that Gliese 445 is approaching the Sun so quickly was the news to me!

Posted in Astronomy | 9 Comments