Quiz Question

Number 188 of a series

Suppose you have a bar of metal, and you know it weighs more than 15 pound and less than 20 pounds. You have three scales (see above), and each has a maximum capacity of 5 pounds. Can you weigh the bar using the three scales? Hint: you can’t cut the bar into pieces.

Post your answer in the comments section below.

Update and Solution

Greg has already provided a workable solution. See the comment below. Here is mine. See the diagram.

The rules don’t say anything about employing extra equipment, so I have added a support block on the left and two wedge-shaped bars. I line the three scales one behind the other on the right and bridge them with a wedge. I place the other wedge on the support and lay the bar to be weighed as shown. Make sure the same amount of the bar overlaps on each end, and now the three scales support half the weight of the bar. Add the readings of the three scales to get half the weight of the bar. Account for the extra weight of the wedge.

If the bar is not uniform in thickness, then reverse the bar and do a second weighing. Sum all six scale measurements, accounting for the weight of the wedge, to get the weight of the bar.

Quiz Question

One of a continuing series

Airliner crosses Vineland Avenue North Hollywood while landing at Burbank Airport

This is the kind of question that sometimes comes up in engineering.

An engineer is tasked with designing a new airplane, with two requirements. The airplane must be able to carry 8 passengers, and the design must use engine model A3-28, made by the boss’s company.

The engineer completes a design, and then does some calculations. The plane will not fly with a single engine of that model, but it will fly with two engines. The engineer’s boss tells him to modify the design and use two of the engines.

The engineer knows this is a bad idea.

Why?

Post your answer in the comments section below.

Quiz Question

One of a continuing series

Industry-OilWellPump

There was a previous Quiz Question involving stress concentration in structural members. That got me to thinking about strength of materials. Here is an interesting fact:

First, modulus of elasticity is a measure of the amount of stretch per cross-sectional area for a given load. For example, the modulus of elasticity of structural steel is in the order of 200 GPa. For specially-treated tool steel the modulus is about 207 GPa—not much better. Yet the tool steel will take ten times the load before rupturing. Does this make sense, and why?

Post your responses in the comments section below.

Update and answer

Greg has the right answer. Hardened steel is nor more rigid than structural steel. It just stretches farther before yielding. It’s counter intuitive that something like a hardened tool bit is just as pliable as an untreated steel bar.

Quiz Question

One of a continuing series

TensileTest

The illustration above shows two structural steel rods. The rod on the left is three inches in diameter. It will pull apart at 254,000 pounds. The rod on the left is the same, except the ends are enlarged to five inches diameter. When you pull on the ends of the rod on the right it will pull apart at measurably less than 254,000 pounds.

Why?

Post your answer in the comments section below.

Update and answer

Greg provided the correct answer, which is “stress concentration.” In the configuration at the left the stress is uniformly distributed, but on the right the stress is concentrated along the outer edge. Some background.

Stress is load concentration, mechanically equivalent to pressure. It is, for example, pounds per square inch. Strain is an objects response to stress. Strain is deformation induced by stress. Strain is a dimensionless quantity, such as inches per inch. It is deformation in inches per total length in inches.

It is not stress that results in failure of a mechanical component, but ultimately strain. For a material, structural steel for example, as you exert a force to pull the sample apart it stretches. The effect is more pronounced in objects made of rubber, and the resulting strain of a steel part may not be apparent. A solid steel bar stretches just like a spring when you pull on the ends, usually not enough for you to notice. Unless you pull really hard. As you pull harder, eventually the steel stops acting like a spring and ceases to offer additional resistance the more you pull. While the load stays constant the steel continues to stretch. The steel has entered into the plastic deformation range.

If you continue to pull you will pull the steel bar apart. This is called rupture, and it’s the failure mode in tension for materials like steel.

When you pull on the sample on the right what you might not notice is that the strain in the center of the bar is not as much as the strain around the outer regions. That is because the extra thickness provided by the shoulder keeps the region adjacent to the narrow section from moving in response to the load. In the shoulder region the center is pulled down farther than the outer region. Within the thin section of the rod the outer region adjacent to the shoulder now stretches more. The strain is greater. The sample on the right, in this example, will start to rupture before the load reaches 254,000 pounds. When the outer region starts to rupture the load shifts to the inner region, and the bar pulls apart before the load reaches 254,000 pounds.

What you see in aircraft structures and in other structures that operate near the limits of materials is great care to reduce stress concentration. To the extent possible, sharp concave corners are eliminated. Structural members that neck down to smaller dimensions do so in a gradual manner.

Quiz Question

One of a continuing series

TechnologyWallSwitch

When you want to turn on the stair lights while going up the stairs, and you want to turn them off once you get to the top, you need this bit of minor technology. The lights are controlled by two switches. One switch is at the bottom of the stairs, and the other switch is at the top. The switches are “double throw” type. That means the switch is connected to two circuits, and the switch has two positions. When you toggle the switch it turns off one circuit and turns on the other. If the light was on, then throwing the switch turns the light off. If the light was off, then throwing the switch turns the light on. Neat.

Now suppose you have a three-story house, and you want to turn the stair lights on when starting up, and you want to turn the lights off when you get to the second floor, and also off when you get to the third floor if you are going that far. How are you going to be able to achieve that? Do you need a new kind of switch?

What if you have a 15-story house?

Post your answer in the comments below.

Update

Some comments have been received, and it’s apparent clarification is needed. This Quiz Question involves single-pole, double-throw switches. Here is how a SPDT switch works:

SPDTSwitch

Update and Solution

The two-switch feature can be extended to more than two switches, but the configuration I had in mind does not work. I had to go to Wikipedia for a workable solution. First, here is how the two-switch configuration works.

 

Technology-DoubleThrow-03

Technology-DoubleThrow-04

Switch 1 and Switch 2 are SPDT switches. In the first configuration no power is supplied to the load, which is typically a lamp. In the second configuration power is supplied to the lamp, and it turns on.

It is trivial to extrapolate from this and determine that flipping either switch turns on an off-lamp and turns off an on-lamp.

To extend this idea to more than two switch requires a different kind of switch, but still a mechanical switch. The following are from Wikipedia:

Readers are invited to visit the Wikipedia article, following which they will be able to install multi-way lighting switches for their five-story condo.

Quiz Question

One of a continuing series

Technology-ShotgunShells-01

Steve and Greg, hold off answering this week’s Quiz Question until some others have had a crack at it.

When you go into Home Depot and purchase some electrical wire, you will notice it comes in different sizes. Wire diameter is specified by its gauge. The deal is the smaller the gauge the thicker the wire. You may not also know that a similar gauge system is used for sheet metal. 8 gauge sheet is getting toward boiler plate thickness. 32 gauge is some thin stuff.

When it comes to shotguns, a gauge system is used to specify the bore. The very common 410 shotgun is not a gauge measure. I could be wrong, but I think a 410 has a 0.410-inch bore. The only common shotgun gauges I know are 10-gauge and 12-gauge. Doesn’t matter.

This week’s Quiz Question is, how is a shotgun gauge determined?

Provide your answer in the comments section below. No need to post your answer on Facebook. I always do that when I post the answers.

Quiz Question

One of a continuing series

GarminGPSIII

Here’s something that came up a few years ago. It has to do with GPS technology, but you don’t need to be GPS-savvy to appreciate it. It goes like this.

GPS employs 32 satellites, whose position is at all times precisely known. Each satellite broadcasts to Earth continuously. The signal contains a lot of stuff, but the critical information from each satellite is:

  • I am here.
  • The time is …

A GPS navigation receiver doesn’t need to know the direction the signal is coming from. All that is necessary to determine your position is the preceding information from each of three or four (four is best) satellites.

From the information received, a navigator can determine where all satellites were at the same time. And it can determine how long it took each satellite’s signal to reach the receiver. Knowing the speed of propagation of the radio signal, the receiver can compute the distance to each of three (or four) known points in space and therefore compute its own position in 3-D space. If it knows the speed of propagation of the radio signal.

The problem is the speed of propagation through the atmosphere to the receiver is not constant. It varies due to the presence of free electrons in the atmosphere. There are two solutions to this difficulty. One is to incorporate an atmospheric model into the receiver’s computation, and this is done. It’s called the Klobuchar model, after the person who developed the mode. It’s not very accurate.

For extreme accuracy, the atmospheric delay can be measured directly. To do this, a second transmission channel is incorporated.

The two satellite transmission channels are called L1 and L2. All receivers can use L1. L2 is encrypted. You have to have a secret key, available only to the U.S. government, to use the L2 channel. The two channels operate on somewhat different frequencies, and the atmosphere delays each channel differently.

And that’s all you need to know if you can receive both L1 and L2. You do not need to know in advance how the atmosphere delays each channel. The receiver can deduce the atmospheric delay from each satellite, and from that it can compute the position of the receiver to very high degree of accuracy.

When I first encountered this it was obvious to me there was not enough information to compute the atmospheric delay. So I asked a guy working on the project how this was supposed to work, and he stopped what he was doing and explained it to me. I still didn’t understand it, but I took my notes back to my cube and looked at it some more. It was an “oh shit” moment. “Of course, dummy.”

And that’s this week’s quiz question. How can a GPS receiver compute the atmospheric delay from the information given, using L1 and L2?

Post your answer as a comment below. I’m going to give this a few days and then post a hint.

Update and hint

I’ve had no activity on this Quiz Question all week. It’s time to provide a hint. Look at the problem again.

You have two radio signals originating from the same location at the same time and arriving at the receiver at different times. Because of ionospheric delay, you don’t know how long it took either signal to traverse the unknown distance to the transmitter. How can you use the information available to determine the distance to the transmitter. Here is the hint.

The satellite is moving. In the order of miles per second. Its distance from the receiver changes from one transmission to the next. How can you use multiple measurements to compensate for the ionospheric delay?

Update and solution

Time’s up. I need to post the solution to last week’s Quiz Question, because tomorrow’s question is going to be related to compensating for ionospheric delay. I’m not going to do the math. Instead, I’m going to pose the question in a different way that will make the solution obvious. It goes like this:

Forget satellites. There are two rail lines running parallel for miles over the horizon. You’re standing at a point along the rail lines waiting for two trains (call them A and B) to arrive. The two trains are going to start at the same time from the same location, and they are going to come at you at different speeds. You don’t know what the speeds are, but you know the speeds of A and B are different, and they are constant.

Trains A and B arrive. A arrives shortly before B. You note the time difference. You don’t have enough information to determine how far away the station (starting point) is. You move down the line a few miles, carefully noting how far you move.

Two more trains, also labeled A and B head your way from the station. Same as before. A arrives, then B arrives later. You record the time difference. Of course the difference is greater, because you are farther from the station, so the trains have had longer to diverge.

Now you ask yourself this question. “How far do I have to walk toward the direction the trains came from for the difference to be zero?” That’s the distance to the station.

In the case of GPS with channels L1 and L2 it’s the satellite that moves, and since the satellite is always telling you where it is, you know how much farther (or closer) it has moved from you between two measurements.

There is obviously more to it, so if anybody still has questions, post a comment and extend the dialogue.

Adventures In Hyperspace

Examine the following:

x-y-coordinates

That’s pretty dull. It is something familiar to any school child, even those not majoring in math or science. The horizontal line is the x-axis of a coordinate system, and the vertical line is the y-axis. They are said to be orthogonal. They are orthogonal in two senses of the word: they form a right angle, and they are mathematically orthogonal. In the second sense measurements along the x-axis are independent of measurements along the y-axis. You can move all you want in the x-direction without affecting your position in the y-direction.

There’s more. For example, suppose x and y represent east and north. Then up can be represented by z. Now we have a system of three coordinates, and they are mutually orthogonal. Movement in the up direction does not affect your position in the east and north directions.

You may need to close your eyes to imagine the next part, but you can add a fourth orthogonal coordinate to this representation of space, and you will have a four-dimensional space. I can’t draw it, so I have to deal with it in a purely mathematical manner. The principles of geometry that hold for three-dimensional space can be extended to higher-dimensional spaces. And that’s what I did.

Next look at this image.

parabola-01-small

The curve comes down from the left, reaches a low point and then continues upward to the right. The task is to locate the low point of the curve. You want to do it by an automated process for reasons that will be made clear later. The straight line is tangent to the curve as shown, and the short vertical line is the x-coordinate at the point of tangency. If you have a mathematical function for the line you can compute the derivative of the function, and the derivative is the slope of the tangent line. You can see that the slope of the tangent line will be zero at the low point, so if you find the point at which the slope is zero, then you have determined, mathematically, the x-coordinate of the low point—the minimum value of the function.

See the diagram.

parabola-03-small

The problem of finding the minimum point of a function devolves into finding the point at which another function, the derivative, is zero. There is a mechanical process for finding the zero of a function, and it’s called the Newton-Raphson method, after Isaac Newton and Joseph Raphson. The utility of this method is it can be performed by a computer.

Now suppose that what you have is not a line function but a surface function. My Microsoft Excel was not able to draw such a surface, so I obtained this one from Wikipedia.

2000px-Paraboloid_of_Revolution

You can find the low point (minimum) of this surface by a method similar to the one above. And that’s what I had to do. There was a project I worked on, no government secrets involved, and we were flying a Convair transport plane carrying multiple imaging systems. We collected visible, infra-red, and laser radar images, and when we got them all back to the lab we needed to determine what the system was looking at at for each image. To do this we needed to know where the airplane was at the time the images were made. As accurately as possible.

We had multiple sources of information. We had an inertial guidance system (INS) and a transponder locater system. And we had more. The problem was the sources of information were in disagreement about where the airplane was with respect to time. It was going to be necessary to apply corrections to the various inputs to minimize the error.

Without getting into the details of how I spent three months of my life, I developed an error function based on the data inputs just mentioned, and my goal was to apply corrections to the input data in order to minimize the error function. See where this fits in? I needed to compute the minimum of a function.

But this was a function in multiple dimensions. Ultimately it grew to a function in ten dimensions. I needed to minimize a function in ten dimensions.

Inspiration came from the surface function. You can use the Newton method to compute the minimum of a surface if you follow this approach, and if you have a well-behaved function.

Start with one axis, for example the x-axis. Compute the minimum for a given value of the y-axis. That is, move along the x-direction while holding y constant. Compute the minimum of the curve defined by the x-values alone. Once you have done that, switch to computing the minimum along the y-axis, using the x-value for the minimum just computed. Repeat until you have finally reached the minimum of the surface.

What I figured would work was to just continue the process into the higher dimensions. Take each axis in turn, holding the other coordinates constant, and compute the minimum. Then go to the next dimension and the next and the next until you have gone through all the dimensions of the error function. Repeat until the solution does not get any better. You have computed the minimum of a function in multiple dimensions.

There was one small hitch. I did not have math functions for the data, just values from a table. To make this computable I needed to supply math functions, and to get math functions I needed to fit curves to the data points. And that’s another story, which I will delve into in another post.

We ran this problem on a VAX computer, which was no slouch of a system in those days, but which would be in the shade of the laptop computer I’m now using to describe all this. And the VAX chugged through solution after solution and computed the position on the ground the cameras were looking at, and when we pulled up the images associated with well-surveyed points, there were the objects that were supposed to be there. To a certain degree of accuracy. It was never perfect. But that’s the way it is in engineering.

Wait, there’s more. After I did all this I learned of an even better approach. It was in a book I already had on my shelf but had not read from cover to cover. It’s a method by Nelder and Mead, and code is available from the book:

We give in §10.4 a somewhat neglected downhill simplex method due to Nelder and Mead. (This use of the word “simplex” is not to be confused with the simplex method of linear programming This method just crawls downhill in a straightforward fashio that makes almost no special assumptions about your function This can be extremely slow, but it can also, in some cases, be ex­tremely robust. Not to be overlooked is the fact that the code is concise and completely self-contained: a general N-dimensional minimization program in under 100 program lines! This method is most useful when the minimization calculation is only an inci­dental part of your overall problem. The storage requirement is of order N2, and derivative calculations are not required.

[Press, William H., et al: Numerical Recipes in C, Cambridge University Press. 1988 p 292]

Quiz Question

One of a continuing series

One of the last useful things I did during my Navy Reserve tour was work in the Sidewinder shop at NAS Dallas. My first encounter with a Sidewinder was educational. There was one on a work stand, and I looked it over. One thing I noticed was that each of the rear stabilizing fins featured a curious mechanism. See the photos. I have scrounged up three so make sure people can get a good view of what I’m talking about.

On the trailing edge of each rear fin, at the outer corner, is a hinged fixture. The fixture is a flat plate with a solid brass disk enclosed in the plate so that the outer edge of the disk is exposed to the wind stream. The edge of the disk is serrated, and the disk is mounted so it can spin freely on its axis. The air stream spins the brass disk at high speeds. The flat plate is attached to the fin along its leading edge by a hinge so it can freely swing to either side.

Sidewinder-01

An AIM-9 Sidewinder Missile installed on an F-14 Tomcat. The AIM-9 is a short range, heat seeking air-to-air missile.

An AIM-9 Sidewinder Missile installed on an F-14 Tomcat. The AIM-9 is a short range, heat seeking air-to-air missile.

Sidewinder-03

So, when the missile is fired, the brass disk is spinning very rapidly, and it’s mounted in the flat plate, which can swing from side to side. And here is the Quiz Question.

What does this arrangement do? If you are an engineer or a physicist looking at this you will figure it out immediately. Engineers and physicists are invited to have a go at this immediately. Post your answers in the comments section below.

UPDATE AND ANSWER

Jim Medding has provided the correct answer that these tabs are used for roll stabilization. He left it for me to provide the mechanism.

This was over 50 years ago, but at the time I first examined this mechanism I was a full time engineering student and a part time Aviation Ordnanceman in the Navy Reserve. About five seconds, and it dawned on me. The tabs work this way:

The outer edge of each brass disk sticks out into the wind, causing the disk to rotate. Applying some basic principles of physics, If the missile rolls (to the right for example), then each disk will be rotated to the right. Because of the direction it’s spinning, the disk will apply torque on the hinged tab and force it to swing to the right, into the air stream on the right side of the fin it’s mounted on. That will produce an aerodynamic force to resist the roll in that direction. Similarly if the missile rolls to the left. These tabs with their rotting disks are an automatic roll stabilization mechanism with a built in control mechanism and requiring no power from the missile control system, which is located way forward on the missile body, anyhow.

It was so slick, I never forgot about it in all this time, and I tip my hat to the engineer who came up with the concept.

Quiz Question

One of a continuing series

See the diagram below. This is how my thermostat works.

Thermostat76Degrees

When the temperature reaches 76° the cooling comes on. The room starts to cool. See the next.

Thermostat74Degrees

When the temperature reaches 74° the cooling shuts off. The room will now start to warm up, but the cooling will not come back on again until the temperature reaches 76° again. This feature prevents the cooling system from cycling on and off frequently around narrow changes in temperature. The feature is called hysteresis. The Quiz Question for this week is this:

What does this have to do with women?

Post your answer as a comment below. I will update the post with the answer on Friday.

Quiz Question

One of a continuing series

This one may be too easy. Here it goes anyhow.

AtlasRocketLaunch

In 1962 an Atlas rocket launched from Cape Canaveral went off course and had to be destroyed. The problem was traced to a snippet of FORTRAN code. Here it is:

DO 55 J = 1.100

What’s wrong with this code? What was it supposed to do, and what did it actually do?

People who’ve been programming for more than 50 years are not allowed to answer this question. Give the newcomers a cut at it first. I will update this post with the answer if there is no correct response by Wednesday.

Update

Steve provided the correct answer. With a comma this is the beginning of a DO loop. With a period it becomes an assignment. FORTRAN programmers should know that the FORTRAN compiler throws away all spaces. The space character is meant only for human eyes. Here’s how the above code appears to the FORTRAN compiler:

DO55J=1.000

The compiler responds by creating a new variable named DO55J and assigns it a value of 1.1. No loop is created, and the rocket is lost.

Quiz Question

One of a continuing series

Racers aren’t allowed to answer this one.

The item pictured below is a carburettor jet.

Cycle Pro Pilot Jet for Keihin Carb

Cycle Pro Pilot Jet for Keihin Carb

Before just about everything went to fuel injection, there were carburettors. Carburettors supply a mixture of air and fuel to the intake ports of piston engines. They work this way.

On the intake stroke of an engine, the piston moves down, drawing air into the cylinder. Air flows rapidly into the cylinder, passing first through the carburettor. The air stream through the carburettor, because of its forward motion, exerts reduced pressure against the side walls of the port. The carburettor jet connects the air stream to a supply of fuel in a float bowl, a small reservoir of fuel in the carburettor body. The reduced pressure causes fuel in the float bowl to be forced through the opening in the jet and into the air stream. The fuel mixes as a spray with the air stream and is carried into the combustion chamber.

The carburettor jet has a measured opening for the fuel. A larger opening will allow more fuel to flow into the air stream. Here’s the question. A motorcycle racer (for example) arriving at the track in the morning observes it’s a warm and humid day. He opens up his race kit box and selects a set of jets with larger openings and installs them in all the carburettors in his engine.

Why? Why does the racer know he’s going to need a larger jet?

Provide your answer as a comment below. If nobody provides the correct answer by Friday I will post it here.

Update

The answer to the Quiz Question is that warm air and moist are are less dense. The equation shows the relationship between pressure (p) and density (ρ) for incompressible flow. I’m using this, since no compression or expansion of the air takes place inside the carburettor throat.

BernoulliEquation

The velocity of the airflow is v. The remaining terms are g and z, the acceleration of gravity and the altitude, neither of which figure into the carburettor operation, all other things being equal. For a given velocity, p and ρ must be proportional. If ρ goes down, p must go down. When the air is less dense the pressure must go down. You need a larger opening in the jet to allow the proper amount of fuel into the air stream, else the cylinder will burn too lean. The result would be a burned piston.

Stand by for another Quiz Question on Monday. And keep reading.

Quiz Question

One of a continuing series

An automobile engine, as everybody has experienced, can develop a maximum amount of power at the drive shaft. This is to be expected. The chemical energy available from fuel and air is fixed for a given flow rate, and an automobile engine reaches a maximum flow rate.

How about a rocket engine? Amazingly, rocket engines don’t have this limitation. Here’s an illustration:

RocketPower

Exhaust gases exit the rear of a rocket at a velocity v (lower case v). The thrust accelerates the rocket body in the opposite direction. At any instant it has a forward velocity V (capital V). The thrust produced by the escaping exhaust gases is constant, no matter how fast the rocket is moving.

The power produced by the rocket motor is given by:

W = fV

The power, W, produced by the rocket motor is the thrust, f, multiplied by the forward velocity, V.

How can this be? The rate of chemical energy expended is constant, controlled by the binding energies of the reactant components. These should not be dependent on forward velocity. Yet, as the rocket goes faster more power is produced. Is this a paradox? If so, what’s the resolution of the paradox?

Post your answer in the comments section below. I will provide the resolution by Friday if nobody posts a solution by then.

Solution

Time’s up. Nobody has been addressing this week’s Quiz Question recently, so I’m posting the answer.

This Quiz Question is a good illustration of a property of energy, kinetic energy in particular. Kinetic energy is dependent on the frame of reference. Revisit the illustration above. In a particular inertial frame of reference the rocket has zero kinetic energy, even several seconds after firing the propellant charge. At any instant in any particular inertial frame of reference the chemical energy of the fuel is being converted to kinetic energy: kinetic energy of the rocket and kinetic energy of the exhaust. At any instant the energy conversion is divided between the rocket and the exhaust. Here are two examples. The frame of reference is the one in which the rocket has zero velocity at the time the propellant charge is fired:

  • At the moment the propellant charge is fired the rocket is not moving. All the chemical energy from the rocket fuel is feeding into the exhaust gases. The rocket motor is developing zero power.
  • Some time later, the rocket is moving. Some of the energy conversion is going into the rocket, and the remainder is going into the exhaust gases. The total is constant. It is the rate of conversion of chemical energy. The rate at which kinetic energy is added to the rocket is the amount of work done on the rocket. This is the definition of the power of the rocket motor. It is the thrust of the rocket motor multiplied by the velocity of the rocket in the frame of reference defined above.

If  any of these points need additional explanation, paste a comment below.

Next Monday, a new Quiz Question. Some people are going to know the answer immediately.

Quiz Question

Forty-five years ago I worked for an engineering consulting company. We did design and development for companies that did not have their own engineers.

This was at the dawn of self-service gas stations, and new companies were getting into the business. One had a chain of stations called Fill-Em-Fast. It was Autotronics out of Houston. What they wanted was to get sales information on a daily basis. This was before the days of the Internet and instant sales reports. What they wanted us to do was to electronically record sales information, gallons and dollars, from each pump. At night, when phone rates were low, the home office would phone each station in turn and retrieve the sales information from the station’s box. We designed that.

This Quiz Question regards a problem involving the retrieval of sales information from pumps. These were the mechanical pumps of your grandfather’s day. There was a mechanical display operated by gears as the fuel was dispensed. We figured all we needed to do was to hook into the rotation of the gear shafts and read the revolutions, or increments of same. That’s where the following image comes in.

gear

We would fit an extra spur gear on the shaft we wanted to read off and just count the gear teeth as they went by. Don Ninke was the EE in charge of this, and he came up with a design that counted teeth with an optical interrupter. An LED would send out light, and a photo-diode would receive it. Gear teeth would interrupt the beam, and we would count the pulses.

It didn’t work. It counted too many teeth. And that’s the Quiz Question. Why didn’t this work? Why was this design counting too many teeth?

Part two of the Quiz Question: How did Don fix the problem and count the correct number of teeth?

As always, post your answers in the comment box below. I will provide the answer Saturday, or before.

Solution

Mike has provided the answer:

Backlash. If you have two sensors, you can tell the direction of motion.
Used this technique on the automobile odometer cable for a time-speed-distance car rally computer.

Mike obviously has had some engineering experience. This is the solution Don Ninke came up with. The applicable term is “quadrature.” Set the two sensors out of phase by 90°. If the gear backs up a tooth, one is subtracted from the count. See the diagram.

Quadrature

Quadrature

 

Counter rotation of the gear shaft typically occurs when the mechanism stops. There’s not much problem when the pump is running, and the shaft is turning continuously. Discounting spurious pulses has two benefits:

  • You don’t over charge the customer for an extra tooth’s worth of gas.
  • At the end of the day you don’t have have the accumulated error of these extra counts, which would screw up the accounting. There were separate shafts for the dispensed gas and dollars charged. They really need to match at the end of the day.

Interesting thing to note: The only computer used in the system was a Data General Nova 1200 mini-computer. That was installed in the home office in Houston, and Wayne van Citters and I did the software for the Nova.

The electronic box that Don Ninke designed employed no processors. It was all series 7400 logic chips from Texas Instruments. It counted pulses and recorded sales from the pumps, accepted calls over a phone line, and spewed out the sales data for the day over an RS-232 link.

About this time (early 1972) we learned there was a company developing a computer on a chip. We requested and obtained a spec sheet dated late 1971 from Intel. I still have it—two Xeroxed sheets on the 8008 microprocessor. It was an amazing time.

Quiz Question

One of a continuing series

See the diagram below.

For about eight years I worked for a company that made document processing systems. One of the company’s inventions was a high-capacity ink jet printer. It worked as shown.

InkJetPrinter

There was a nozzle that spit out 105,000 ink drops per second. A variable voltage was applied to the nozzle, which charge was retained by each ink drop as it exited the nozzle. The voltage was changed for each drop.

The drops passed between two charged plates. Drops with differing charges were deflected in different directions by an amount proportional to their charge. It was a touchy situation. The voltage needed to be something like 5000 volts between the two plates. Not enough voltage, and the drops would not be deflected enough. Too much voltage, and electricity would arc between the plates. I spent weeks getting a configuration that would work. Ink drops were properly deflected, and there was no arcing.

We installed our systems in major banking and financial centers around the world. Then I asked a question. “Do we have any customers in Denver?”

And the quiz question is this. Why did I ask that question?

Solution

Mike apparently saw the answer right away. I asked him to hold off before posting his comment. And here it is.

From high school physics we know that dry air breaks down at 25,000 Volts per inch. That’s dry air at sea level. When the air gets thinner at higher altitudes it breaks down at lower voltages. This arrangement featured two charged plates operating on the verge of breakdown. That was one of the big problems, as noted in the statement of the problem.

It was obvious when we finally got this thing to working that it wouldn’t take much to tip the balance and induce arcing. An increase of one mile in altitude would surely do it.

As it turns out we didn’t have any systems installed in Denver, La Paz or any other troublesome locations.

Provide your responses in the comments below.

Quiz Question

One of a continuing series

Here is a photo from a nearby service station. Note the price of diesel.

PriceOfDiesel

The first question is:

1. Why does diesel fuel cost more? That is to say, why are drivers, especially big truckers, willing to pay more?

The second question is for after you answer the first question. It will be revealed after you answer the first question:

2. Why?

Solution

There have a few responses, at least one that accounted for some of diesel’s advantage. Prasad commented:

Diesel is denser and longer hydro-carbon chains than gasoline, which is lighter and more volatile. Hence gasoline has less energy per gallon than a gallon of diesel.

Since it can make cars/trucks go farther with the same volume of fuel it is still economical to buy diesel at a higher price than gasoline (for the same volume.)

I guess they are charging it by energy content.

Here are the facts concerning relative energy content:

As of 2010, the density of petroleum diesel is about 0.832 kg/L (6.943 lb/US gal), about 11.6% more than ethanol-free petrol (gasoline), which has a density of about 0.745 kg/L (6.217 lb/US gal). About 86.1% of the fuel mass is carbon, and when burned, it offers a net heating value of 43.1 MJ/kg as opposed to 43.2 MJ/kg for gasoline. However, due to the higher density, diesel offers a higher volumetric energy density at 35.86 MJ/L (128,700 BTU/US gal) vs. 32.18 MJ/L (115,500 BTU/US gal) for gasoline, some 11% higher, which should be considered when comparing the fuel efficiency by volume.

That’s an 11% advantage over gasoline. Since the fuels are priced by volume this gives diesel a slight advantage with the prices shown in the photograph, which shows diesel costing 10% more. In order to get a price advantage that would make it worth while to drive diesel over gasoline, some other considerations need to be taken into account. These would involve the way diesel engines operate compared to gasoline engines.

Gasoline engines throttle the intake air. Diesel engines do not. A diesel engine would not work if you throttled the air intake. Operation of a diesel engine requires a full cylinder of intake air on the compression stroke in order to generate the ignition temperature. Diesel engines don’t use spark plugs. Each cylinder has a glow plug to ignite the fuel when starting the engine. After that the engine relies on injecting fuel into the hot, compressed air in the cylinder.

Diesel engines run at leaner fuel mixtures when the pedal is not pressed all the way down. The throttle is a fuel throttle, not an air throttle. Backing off on the throttle decreases the fuel injected into the cylinder, but not the air.

If you run a gasoline engine lean the combustion temperature goes up, resulting in melted pistons and burned valves. Not so with a diesel engine. Leaning the mixture in the cylinder only decreases the amount of fuel in the combustion and does not raise the combustion temperature.

Also, for comparable reasons, diesel engines typically run at higher compression ratios than gasoline engines. A 12:1 compression ratio is a racing engine for gasoline. Diesel engine compression ratios start at 12:1 and go up from there. Typical would be 14:1. Some engines run at 18:1. Basic principles of thermodynamics produce better fuel economy (more mechanical work from comparable chemical energy) for higher compression ratios.

Diesel vehicles may be making a comeback. Diesel engines are more fuel-efficient and have more low-end torque than similar-sized gasoline engines, and diesel fuel contains roughly 10% to 15% more energy than gasoline. So, diesel vehicles can often go about 20% to 35% farther on a gallon of fuel than their gasoline counterparts. Plus, today’s diesel vehicles are much improved over diesels of the past.

Given this, the 10% hit on fuel price does not look so bad.

Quiz Question

One of a continuing series

This one really is a quiz question I encountered in engineering school. We had this class on Engineering Design, and it was being taught by a Ph.D. candidate. I know his name, but I won’t reveal it. He thought we needed some stimulation, so he popped a quiz on us. “Boys and girls, here is a design problem for you.” He showed us something like the drawing below.

PipesAnndValves

What you see here are two pipes running side by side, with valves. This is a schematic drawing. The pipes and valves do not look like this in real life. Those things with the Xs represent the valves. The problem, our professor told us, is you do not want both valves to be open at the same time. One valve is on, the other must be off. If you want to turn on the off valve, then you have to first turn off the on valve.

We are to design a mechanism that ensures we can perform this operation. The mechanism must ensure that it is impossible for both valves to be on at the same time.

There is an interesting story behind this, and I will tell it to you as a hint. Meaning, the solution was simple. I looked at this for about five seconds and said to myself, “Of course!” Then I drew the solution on my test paper and turned it in. I was then allowed to go about my business.

Came the next class period the professor returned our test papers, and I only received a B. How come, I asked. I provided the correct answer. How come I only got a B? The professor told me that I had obviously seen the problem before, so he could not give me full credit. This was one of those WTF moments I have experienced in college (and other times). I got an A.

Provide your answer as a comment below. You will likely need to do a drawing, so send a sketch by email. I will attach your sketch to the post. This one should be easy, so I will provide my solution on Saturday.

UPDATE

I’ve changed my mind. I’m going to post the answer now. People have submitted a number of worthy solutions. Here is one submitted by Michael Brown:

BrownSolution

Nobody has come up with the optimum solution. I submit here a re-creation of my solution from 50 years ago:

Interlock

Here’s how it works. Each valve stem has attached to it a disk. Each disk has a cutout that has the same radius as the disks. The valve stem on the left is constrained by stops (not shown) that limit its travel from the position shown to 90 degrees CCW. The disk on the right is constrained from its position shown to an orientation 90 degrees CCW. When the cutout is in the upper position, the valve is on. When the cutout is directed to the side, the valve is completely off.

I call this the optimum solution because no other moving parts are required besides the valve stems with their attached disks.

As I mentioned, this problem was given to us as an in-class quiz with no preparation. I had a solution submitted within 5 minutes. I had not seen this solution prior to the class period, but I had the advantage over my classmates. I had studied the mechanisms associated with motions picture film transport. I was familiar with the Geneva drive. Here it is:

GenevaDrive

 

The disk on the right turns. The star wheel is constrained from rotating by the machined shoulder on the disk. When the pin attached to the disk engages one of the slots, the shoulder cutout allows the pin to rotate the star wheel. The result is the star wheel shaft alternately rotates, stops for a time, then rotates again. This is what is needed to advance motion picture film during recording and during projection. This is what came to mind when I considered the problem of the two valves. The solution flowed from there. And it got me a B on the quiz.

Quiz Question

Base64

This came up when I was working for a company that did secure communications. We were doing the software that formatted the messages and sent them out over the secure link. The messages had to be encrypted prior to transmission. To save communication bandwidth we also wanted to compress the messages.

The question came up: Do we want to encrypt the messages and then compress them, or do we want to compress the messages and then encrypt them. The answer was obvious.

This is a question in two parts:

1. Which should you do first, encrypt or compress?

2. Why?

As with all of these, responses should be posted as a comment to this post. The answer will be provided in slightly more than a week.

Time’s Up!

I have had a couple of responses, and none of them have given the problem any serious thought. I also advised readers to try these concepts out. A small bit of experimentation would have revealed the what, but not maybe the why. And what is the answer?

1. Compress first, then encrypt.

2. You must compress first, because an encrypted file cannot be compressed.

A few minutes ago I ran this simple demonstration. I generated a text file—the first 12 chapters of Genesis. That seemed appropriate.

Next I constructed a simple encryption program, and encrypted the first 12 chapters of Genesis.

Now I had two files—the first 12 chapters of Genesis and the first 12 chapters of Genesis encrypted.

Then I zipped both files. Here is the result:

genesis.txt    39 KB
outfile.txt    39 KB (encrypted Genesis)
genesis.zip    12 KB
outfile.zip    39 KB

The encrypted file will not compress.

People who read the blog would have known this already. This principle was previously explained two years ago:

https://skeptic78240.wordpress.com/2012/08/06/the-codebreakers/

This is the thing that came to me 20 years ago when the question was first posed. It was like, “Yeah, dummy. If you have an encrypted message, and you can compress it, then you are a long way down the road to decoding it.” Since decoding encrypted messages is not usually trivial, compression is out of the question.

 

Quiz Question

These quiz questions are not meant to be hard. This may be one of the easiest.

Three highly-paid professionals were discussing a design solution. It involved maximizing the efficiency of a packet router. See the diagram below.

RouterTraffic

Packets come into a router by multiple data links, shown by the arrows on the left. The router analyzes each packet and determines which link to send it out on. The output links are shown by the arrows on the right. This is a schematic representation, of course.

The router has the ability to store packets temporarily. Processing of a packet does not begin until all of the packet is received and stored. Packets may not be sent out immediately. If the output link is busy its packets are queued until the link is available.

The question came up. The analysis was something like this:

  • 500,000 packets were coming in per second.
  • The average time a packet spent transiting the router was 10 milliseconds (0.010 seconds).

The question was, how many packets are stored in the router at any one time?

Two of us saw the answer immediately. The third person required some explaining. What is the answer?

Solution

This has gone on long enough. Nobody is submitting any sensible answers, so here is the solution:

It’s simple: 500,000 packets per second, average time in the router is 0.010 seconds: 500,000 × 0.010 = 5000 packets in the router at any one time, on average.

This simple principle can be applied to all kinds of every day life situations. How much water is in a channel? How many cars are there on a length of the freeway? How many people are standing in line to buy tickets at the movie theater? It’s from queuing theory 001.

Structural Defficiency

KansasCitySkywalkCollapse-01

From Wikipedia

I started my working career as an engineer right out of college, working for the university from which I had just graduated. And quickly I began to hear engineering horror stories. One was the case of the reversed viewport.

The University of Texas Austin campus had a new accelerator building where they intended to do research in (relatively) high-energy physics. It was a linear accelerator where charged particles were shot down an evacuated tube under the force some very high voltage differences. In order to do something with these high-speed particles you have to insert something in their path inside the evacuated tube, and to do that you have to work with a port, an engineered hole, in the side of the tube. That’s where the problem came in.

The carefully designed port was essentially a hole in the tube wall, and there was a vacuum flange built into the port. The vacuum flange allowed researchers to bolt a matching flange to the port, tighten the bolts, and obtain a vacuum-tight seal.

Came the time the researchers needed to add a new device to the port, and that device needed to have a matching vacuum flange. No problem. They sent over engineering drawings of the vacuum flange to the office where I worked in the Physics building. This was before I came to work, so others told me the story later, and it was funny. Sort of.

The engineers in the physics building designed the required device with the matching flange. Then somebody had the whole business constructed in the machine shop and sent it over to the accelerator lab. It did not fit. The bolt holes did not line up. You wanted to say WTF?

An investigation ensued. Yes, the engineers received the correct drawings of the vacuum flange. Yes, the new device was designed correctly according to the engineering drawings. Yes, the flange was constructed according to the engineering drawings. No, the damn thing did not fit. And then they revealed the source of the problem.

When the accelerator was constructed the machinist who produced the vacuum flange reversed the drawing. He constructed a mirror image of the port design.

No problem. The accelerator lab guys figured they could just make the matching flanges to conform to the as-built. And they did. And everything worked just fine, because it really didn’t matter whether the flange was left-handed or right-handed.

But they never changed the drawings. When they asked the engineers in the physics building to build a new device they never told them about the change. Nobody talked to nobody, and everybody just went along doing their work and drawing their pay.

A few years later lightening struck again. This time I was working for an engineering company, and we got the job to design and build a bracket to hold a gold-coated mirror for the McDonald Observatory. No problem. My boss had done this before. He had the drawings from the previous bracket he had designed. He gave me the drawings and told me to make a new design, only use a 45-degree angle.

I did that, and we had it built, and we shipped the finished product out to the observatory. There was a tight schedule. The astronomer had reserved only a small window of time on the telescope. The astronomer came out from Langley to the observatory in West Texas. We got a phone call. The damn thing did not fit.

My boss was non-plussed. What was the problem? We built the first one, and there was no problem. There should be no problem with the new one.

Actually, there was a problem with the first one. Only, at the observatory the engineers had spotted the problem immediately and had their machinists produce a correct version of the bracket. And they never told anybody. They didn’t tell the NASA guy from Langley when he came out the first time to use the telescope, and they never told my boss. And we spent government research bucks building yet another wrong bracket. And it was not cheap.

And this kind of stuff can be deadly:

The Hyatt Regency hotel walkway collapse occurred at the Hyatt Regency Kansas City in Kansas City, Missouri, United States on Friday, July 17, 1981. Two vertically contiguous walkways collapsed onto a tea dance being held in the hotel’s lobby. The falling walkways killed 114 and injured a further 216 people. At the time, it was the deadliest structural collapse in U.S. history, not surpassed until the collapse of the south tower of the World Trade Center in 2001.

This was another of those situations. The original designer, Gillum-Colaco International Inc., produced a workable design. You may need to visualize this:

  • There was a second-floor walkway, a third-floor walkway and a fourth-floor walkway, all suspended from the lobby ceiling structure by steel rods.
  • The forth-floor walkway was suspended directly by steel rods attached to the building structure above.
  • The second-floor walkway was suspended directly below the fourth-floor walkway, on the same rods. See the drawing.

kc_walk_proposed

The rods attached to the upper building structure ran straight down, passing through the supports for the fourth-floor walkway to the second-floor walkway. The rods were designed to hold the load of both walkways. The fourth-floor walkway was supported by the rods by threaded nuts, seen in the drawing. To install the nuts, the rods had to be threaded their length up to the fourth-floor walkway.

First some notes about the strength of threaded fasteners. An SAE (Society of Automotive Engineers) machine screw will, under tensile load, pull in two before the threads will strip, provided at least six threads are engaged. If the nut engages at least six threads there would be no danger of the threads stripping under any load the screw will sustain. Threading a load-bearing bar, like a screw, produces some weaknesses over an unthreaded bar. Strengths of threaded fasteners are ultimately determined by putting samples in a testing machine and pulling until the rod separates.

The problem came when the builder, Havens Steel Company, did not want to thread all this much steel rod. They proposed an alternative.

kc_walk_actual

Now Havens needed to thread only a few inches of steel rod. Everything should work the same, right? No. Look at the drawing at the top of the page, and follow along with this explanation:

Due to the addition of another rod, the load on the nut connecting the fourth floor segment was increased. The original load for each hangar rod was to be 90kN, but the alteration increased the load to 181kN. The box beams were welded horizontally and therefore could not hold the weight of two walkways. During the collapse, the box beam split and the bottom rod pulled through the box beam resulting in the collapse.

For non-engineers out there, 90kN is 90,000 Newtons, about 20,183 pounds. Multiple rods supported each section of the walkway, so the design was safe for the expected load.

Look at the drawing at the top of the page again. There is something else wrong with the altered design. Offsetting the down rod and the up rod produces a shear force on the beam that was not intended by the original designers. GCI, Inc. surrendered their professional engineering expertise to Havens, a company lacking any such credentials. And people died.

All this came to mind a few days ago when a former college roommate sent me this photo from Jakarta.

IndonesianArchitecture

Here is what Byron Black had to say in his note:

Daddy makes enquiries. An architect friend hollers out laughing down the telephone line. “Oh that place! They hired these whiz-bang architects from Paris; the geniuses decided they’d go for a supple, light structure. Cut back on the reinforcing rods.
“Then the owner changed her cute little mind and added fifteen more floors to the original design. Wanted to sell more units.
“Nobody thought to tell these guys that the buildings they’d drawn up, and the precise angle they’d set them at, were going to pick up the wind and sail back and forth. Even when the contractor was building them, the workers all bundled up and scared shitless from the swaying, the architects never found out.
The owner added 15 additional floors. And never told the architects. And never changed the design to strengthen the lower structure to support the additional load.
I’m thinking there must be a Take An Engineer Out To Lunch Day sometime soon. Do that. And have a good talk.