The ‘all-in-a-row’ circuit diagram convention for series and parallel circuits

Circuit diagrams can be seen either as pictures or abstractions but it is clear that pupils often find it hard to recognise the circuits in the practical situation of real equipment. Moreover, Caillot found that students retain from their work with diagrams strong images rather than the principles they are intended to establish. The topological arrangement of a diagram or a drawing presents problems for pupils which are easily overlooked. It seems that pupils’ spatial abilities affect their use of circuit diagrams: they sometimes do not regard as identical several circuits, which, though identical, have been rotated so as to have a different spatial arrangement. […] Niedderer found that pupils, when asked whether a circuit diagram would ‘work’ in practice, more often judged symmetrical diagrams to be functioning than non-symmetric ones.

Driver et al. (1994): 124 [Emphases added]

For the reasons outlined by Driver and others above, I think it’s a good idea to vary the way that we present circuit diagrams to students when teaching electric circuits. If students always see circuit diagrams presented so that (say) the cell is at the ‘top’ and ‘facing’ a certain way; or that they are drawn so that they are symmetrical (which is an aesthetic rather that a scientific choice), then they may well incorrectly infer that these and other ‘accidental’ features of our circuit diagrams are the essential aspects that they should pay the most attention to.

One ‘shake it up’ strategy is to redraw a circuit diagram using the ‘all-in-a-row’ convention.

If you arrange the real components in the ‘all-in-a-row’ arrangement, then a standard digital voltmeter has, what is in my opinion a regrettably underused functionality, that will show:

  • ‘positive’ potential differences: that is to say, the energy added to the coulombs as they pass through a cell or the electromotive force; and
  • ‘negative’ potential differences: that is to say, the energy removed from each coulomb as they pass through a resistor; these can be usefully referred to as ‘potential drops’

This can be shown on circuit diagrams as shown below/

In other words, the difference between the potential difference across the cell (energy being transferred into the circuit from the chemical energy store of the cell) is explicitly distinguished from the potential difference across the resistor (energy being transferred from the resistor into the thermal energy store of the surroundings). The all-in-a-row convention neatly sidesteps a common misconception that the potential difference across a cell is equal to the potential difference across a resistor: they are not. While they may be numerically equal, they are different in sign, as a consequence of Kirchoff’s Second Law. As I have suggested before, I think that this misconception is due to the ‘hidden rotation‘ built into standard circuit diagrams.

Potential divider circuits and the all-in-a-row convention

Although I am normally a strong proponent of the ‘parallel first heresy‘, I’ll go with the flow of ‘series circuit first’ in this post.

Diagrams 2 and 3 in the sequence show that the energy supplied to the coulombs (+1.5 V or 1.5 joules per coulomb) by the cell is transferred from the coulombs as they pass through the double resistor combination. Assuming that R1 = R2 then, as diagram 4 shows, 0.75 joules will be transferred out of each coulomb as they pass through R1; as diagram 5 shows, 0.75 joules will be transferred out of each coulomb as they pass through R2.

Parallel circuits and the all-in-a-row convention

I’ve written about using the all-in-a-row convention to help explain current flow in parallel circuits here, so I will focus on understanding potential difference in parallel circuit in this post.

Again, diagrams 2 and 3 in the sequence show that the positive 3.0 V potential difference supplied by the cell is numerically equal (but opposite in sign) to the negative 3.0 V potential drop across the double resistor combination. It is worth bearing in mind that each coulomb passing through the cell gains 3.0 joules of energy from the chemical energy store of the cell. Diagrams 4 and 5 show that each coulomb passing through either R1 or R1 loses its entire 3.0 joules of energy as it passes through that resistor. The all-in-a-row convention is useful, I think, for showing that each coulomb passes through just one resistor as it makes a single journey around the circuit.

Acknowledgements

Circuit simulations from the excellent https://phet.colorado.edu/sims/html/circuit-construction-kit-dc/latest/circuit-construction-kit-dc_en.html

Circuit diagrams drawn using https://www.circuit-diagram.org/editor/

Reference

Driver, R., Squires, A., Rushworth, P., & Wood-Robinson, V. (1994). Making sense of secondary science: Research into children’s ideas. Routledge.

Just a moment . . .

Many years ago, I was taught this compact and intuitive convention to show turning moments. I think it should be more widely known, as it not only is concise and powerful, but also meets the criterion of being an effective form of dual coding which is helpful for both GCSE and A-level Physics students.

Let’s look at an example question.

Let’s start by ‘annotating the hell’ out of the diagram.

We could take moments around any of the marked points A-E on the diagram. However, we’re going to take moments around B as it enables us to ignore the upward reaction force acting on the rule at B. (This force is not shown on the diagram.)

To indicate that we’re going to be considering the sum of the clockwise moments about point B, we use this intuitive notation:

If we consider the sum of anticlockwise moments about point B, we use this:

We lay out our calculations of the total clockwise and anticlockwise moments about B as follows.

We show that we are going to apply the Principle of Moments (the sum of clockwise moments is equal to the sum of anticlockwise moments for an object in equilibrium) like this:

The rest, as they say, is not history but algebra:

I hope you find this ‘momentary’ convention useful(!)

Split ring commutator? More like split ring commuHATER!

Students find learning about electric motors difficult because:

  1. They find it hard to predict the direction of the force produced on a conductor in a magnetic field, either with or without Fleming’s Left Hand Rule.
  2. They find it hard to understand how a split ring commutator works.

In this post, I want to focus on a suggested teaching sequence for the action of a split ring commutator, since I’ve covered the first point in previous posts.

Who needs a ‘split ring commutator’ anyway?

We all do, if we are going to build electric motors that produce a continuous turning motion.

If we naively connected the ends of a coil to power supply, then the coil would make a partial turn and then lock in place, as shown below. When the coil is in the vertical position, then neither of the Fleming’s Left Hand Rule (FLHR) forces will produce a turning moment around the axis of rotation.

When the coil moves into this vertical position, two things would need to happen in order to keep the coil rotating continuously in the same direction.

  • The current to the coil needs to be stopped at this point, because the FLHR forces acting at this moment would tend to hold the coil stationary in a vertical position. If the current was cut at this time, then the momentum of the moving coil would tend to keep it moving past this ‘sticking point’.
  • The direction of the current needs to be reversed at this point so that we get a downward FLHR force acting on side X and an upward FLHR force acting on side Y. This combination of forces would keep the coil rotating clockwise.

This sounds like a tall order, but a little device known as a split ring commutator can help here.

One (split) ring to rotate them all

The word commutator shares the same root as commute and comes from the Latin commutare (‘com-‘ = all and ‘-mutare‘ = change) and essentially means ‘everything changes’. In the 1840s it was adopted as the name for an apparatus that ‘reverses the direction of electrical current from a battery without changing the arrangement of the conductors’.

In the context of this post, commutator refers to a rotary switch that periodically reverses the current between the coil and the external circuit. This rotary switch takes the form of a conductive ring with two gaps: hence split ring.

Tracking the rotation of a coil through a whole rotation

In this picture below, we show the coil connected to a dc power supply via two ‘brushes’ which rest against the split ring commutator (SRC). Current is flowing towards us through side X of the coil and away from us through side Y of the coil (as shown by the dot and cross 2D version of the diagram. This produces an upward FLHR force on side X and a downward FLHR force on side Y which makes the coil rotate clockwise.

Now let’s look at the coil when it has turned 45 degrees. We note that the SRC has also turned by 45 degrees. However, it is still in contact with the brushes that supply the current. The forces on side X and side Y are as noted before so the coil continues to turn clockwise.

Next, we look at the situation when the coil has turned by another 45 degrees. The coil is now in a vertical position. However, we see that the gaps in the SRC are now opposite the brushes. This means that no current is being supplied to the coil at this point, so there are no FLHR forces acting on sides X and side Y. The coil is free to continue rotating clockwise because of momentum.

Let’s now look at the situation when the coil has rotated a further 45 degrees to the orientation shown below. Note that the side of the SRC connected to X is now touching the brush connected to the positive side of the power supply. This means that current is now flowing away from us through side X (whereas previously it was flowing towards us). The current has reversed direction. This creates a downward FLHR force on side X and an upward FLHR force on side Y (since the current in Y has also reversed direction).

And a short time later when the coil has moved a total of180 degrees from its starting point, we can observe:

And later:

And later still:

And then:

And then eventually we get back to:

Summary

In short, a split ring commutator is a rotary switch in a dc electric motor that reverses the current direction through the coil each half turn to keep it rotating continuously.

A powerpoint of the images used is here:

And a worksheet that students can annotate (and draw the 2D versions of the diagrams!) is here:

I hope that this teaching sequence will allow more students to be comfortable with the concept of a split ring commutator — anything that results in a fewer split ring commuHATERS would be a win for me 😉

Speed of sound using Phyphox

You can get encouragingly accurate values for the speed of sound in the school laboratory using a tape measure and two smartphones (or tablets) running the Phyphox app.

Phyphox (pronounced FEE-fox) is an award-winning free app that was developed by physicists at Aachen University who wanted to give users direct access to the many sensors (e.g. accelerometers and magnetometers) which are standard features on many smartphones. In effect, it turns even the humblest smartphone or tablet into a multifunctional measuring instrument comparable to one of Star Trek’s famous ‘tricorders’.

Star Trek Tricorder (not Phyphox)

To measure the speed of sound, we need two smartphones or tablets running Phyphox. We will be using the ‘Acoustic Stopwatch’ which measures the time between two acoustic events.

Phyphox main screen

Step 1: Place two devices a measured distance s apart. Typically about 2 or 3 m should be OK otherwise the sound made by A will not be loud enough to control stopwatch B — this can be established through trial and error and depends on many factors including the background noise level.

Step 2: Person A makes a loud sound (a clap or a single syllable shout like ‘Hey!’ is good).

Step 3: Person B and stopwatch B wait for sound created by A to reach them.
Step 4: The sound reaches stopwatch B and starts it running and B hears the sound.

Step 5: B makes a loud sound in response.

Step 6: The loud sound made by B reaches stopwatch B and makes it stop. Let’s call the time displayed tB. This measures the delay between the sound from A reaching stopwatch B and B reacting to the sound and stopping the clock. It includes the time taken for the initial sound travelling from the device to B, B’s reaction time, and the time taken for the sound made by B to travel to stopwatch B. B does not have to be particularly ‘quick off the mark’ to respond to A’s sound — although the shorter the time then the less likely it is then a background noise will interrupt the experiment.

Step 7: The sound made by B travels toward stopwatch A.

Step 8: The sound made by B reaches stopwatch A and makes it stop. Let’s call the time recorded on stopwatch A tA.

If we break down the events included in tA and tB, we find that tA is always larger than tB:

If we subtract tAtB we find that this is the time it takes sound to travel a distance of 2s.

Step 9: We can therefore use this formula to find the v the speed of sound.

We have found that this method works well giving mean values of about 350 m/s for the speed of sound (which will vary with air temperature). This video models the method.

And so we have a reasonably practicable method of measuring the speed of that doesn’t involve complex equipment that is unfamiliar to most students; or a method that involves finding a large and featureless wall that produces a detectable echo when a loud sound is made from a point several metres in front of it.

I don’t know about you, but as a physics teacher, I feel cheated. If it doesn’t involve a double beam oscilloscope, a signal generator, two microphones and two power amplifiers then I simply don’t want to know about it . . .

The Rite of AshkEnte, quite simply, summons and binds Death.  Students of the occult will be aware that it can be performed with a simple incantation, three small bits of wood and 4cc of mouse blood, but no wizard worth his pointy hat would dream of doing anything so unimpressive; they knew in their hearts that if a spell didn’t involve big yellow candles, lots of rare incense, circles drawn on the floor with eight different colours of chalk and a few cauldrons around the place then it simply wasn’t worth contemplating.

Terry Pratchett, ‘Mort’ (1987)

Deriving centripetal acceleration

When I was an A-level physics student (many, many years ago, when the world was young LOL) I found the derivation of the centripetal acceleration formula really hard to understand. What follows is a method that I have developed over the years that seems to work well. The PowerPoint is included at the end.

Step 1: consider an object moving on a circular path

Let’s consider an object moving in circular path of radius r at a constant angular speed of ω (omega) radians per second.

The object is moving anticlockwise on the diagram and we show it at two instants which are time t seconds apart. This means that the object has moved an angular distance of ωt radians.

Step 2: consider the linear velocities of the object at these times

The linear velocity is the speed in metres per second and acts at a tangent to the circle, making a right angle with the radius of the circle. We have called the first velocity v1 and the second velocity at the later time v2.

Since the object is moving at a constant angular speed ω and is a fixed radius r from the centre of the circle, the magnitudes of both velocities will be constant and will be given by v = ωr.

Although the magnitude of the linear velocity has not changed, its direction most certainly has. Since acceleration is defined as the change in velocity divided by time, this means that the object has undergone acceleration since velocity is a vector quantity and a change in direction counts as a change, even without a change in magnitude.

Step 3a: Draw a vector diagram of the velocities

We have simply extracted v1 and v2 from the original diagram and placed them nose-to-tail. We have kept their magnitude and direction unchanged during this process.

Step 3b: close the vector diagram to find the resultant

The dark blue arrow is the result of adding v1 and v2. It is not a useful operation in this case because we are interested in the change in velocity not the sum of the velocities, so we will stop there and go back to the drawing board.

Step 3c switch the direction of velocity v1

Since we are interested in the change in velocity, let’s flip the direction of v1 so that it going in the opposite direction. Since it is opposite to v1, we can now call this -v1.

It is preferable to flip v1 rather than v2 since for a change in velocity we typically subtract the initial velocity from the final velocity; that is to say, change in velocity = v2 – v1.

Step 3d: Put the vectors v2 and (-v1) nose-to-tail

Step 3e: close the vector diagram to find the result of adding v2 and (-v1)

The purple arrow shows the result of adding v2 + (-v1); in other words, the purple arrow shows the change in velocity between v1 and v2 due to the change in direction (notwithstanding the fact that the magnitude of both velocities is unchanged).

It is also worth mentioning that that the direction of the purple (v2v1) arrow is in the opposite direction to the radius of the circle: in other words, the change in velocity is directed towards the centre of the circle.

Step 4: Find the angle between v2 and (-v1)

The angle between v2 and (-v1) will be ωt radians.

Step 5: Use the small angle approximation to represent v2-v1 as the arc of a circle

If we assume that ωt is a small angle, then the line representing v2-v1 can be replaced by the arc c of a circle of radius v (where v is the magnitude of the vectors v1 and v2 and v=ωr).

We can then use the familiar relationship that the angle θ (in radians) subtended at the centre of a circle θ = arc length / radius. This lets express the arc length c in terms of ω, t and r.

And finally, we can use the acceleration = change in velocity / time relationship to derive the formula for centripetal acceleration we a = ω2r.


Well, that’s how I would do it. If you would like to use this method or adapt it for your students, then the PowerPoint is attached.

Please Like or leave a comment if you find this useful 🙂

We all adore Caloric

We all adore a Kia-Ora

Advertising slogan for ‘Kia-Ora’ orange drink (c. 1985)

Energy is harder to define than you would think. Nobel laureate Richard Feynman defined ‘energy’ as

a numerical quantity which does not change when something happens. It is not a description of a mechanism, or anything concrete; it is just a strange fact that we can calculate some number and when we finish watching nature go through her tricks and calculate the number again, it is the same. […] It is important to realize that in physics today, we have no knowledge of what energy is. […] It is an abstract thing in that it does not tell us the mechanism or the reasons for the various formulas.

Feynman Lectures on Physics, Vol 1, Lecture 4 Conservation of Energy (1963)

Current secondary school science teaching approaches to energy often picture energy as a ‘quasi-material substance’.

By ‘quasi-material substance’ we mean that ‘energy is like a material substance in how it behaves’ (Fairhurst 2021) and that some of its behaviours can be modelled as, say, an orange liquid (see IoP 2016).

The eight energy stores as suggested by the IoP

And yet, sometimes these well-meaning (and, in my opinion, effective) approaches can draw some dismissive comments from some physicists.

The Simpsons Comic Book Guy character saying "Picturing energy as a quasi-material substance? That teaching approach smacks of the oh-so-discredited 'Caloric' theory of energy to me . . ."
The Simpsons’ Comic Book Guy weighs in the ‘Teaching Energy’ debate

What was the ‘Caloric Theory of Energy’?

To begin with, there was never a ‘Caloric Theory of Energy’ since the concept of energy had not been developed yet; but the Caloric Theory of Heat was an important step along the way.

Caloric was an invisible, weightless and self-repelling fluid that moved from hot objects to cold objects. Antoine Lavoisier (1743-1794) supposed that the total amount of caloric in the universe was constant: in other words, caloric was thought to be a conserved quantity.

Caloric was thought to be a form of ‘subtle matter’ that obeyed physical laws and yet was so attenuated that it was difficult to detect. This seems bizarre to our modern sensibilities and yet Caloric Theory did score some notable successes.

  • Caloric explained how the volume of air changed with temperature. Cold air would absorb caloric and thus expand.
  • The Carnot cycle which describes the maximum efficiency of a heat engine (i.e. a mechanical engine powered by heat) was developed by Sadi Carnot (1796-1832) on the basis of the Caloric Theory

Why Caloric Theory was replaced

It began with Count Rumford in 1798. He published some observations on the manufacturing process of cannons. Cannon barrels had to be drilled or bored out of solid cylinders of metal and this process generated huge quantities of heat. Rumford noted that cannons that had been previously bored produced as much heat as cannons that were being freshly bored for the first time. Caloric Theory suggested that this should not be the case as the older cannons would have lost a great deal of caloric from being previously drilled.

The fact that friction could seemingly generate limitless quantities of caloric strongly suggested that it was not a conserved quantity.

We now understand from the work James Prescott Joule (1818-1889) and Rudolf Clausius (1822-1888) that Caloric Theory had only a part of the big picture: it is energy that is the conserved quantity, not caloric or heat.

As Feynman puts it:

At the time when Carnot lived, the first law of thermodynamics, the conservation of energy, was not known. Carnot’s arguments [using the Caloric Theory] were so carefully drawn, however, that they are valid even though the first law was not known in his time!

Feynman Lectures on Physics, Vol 1, Lecture 44 The Laws of Thermodynamics

In other words, the Caloric Theory is not automatically wrong in all respects — provided, that is, it is combined with the principle of conservation of energy, so that energy in general is conserved, and not just the energy associated with heat.

We now know, of course, that heat is not a form of attenuated ‘subtle matter’ but rather the detectable, cumulative result of the motion of quadrillions of microscopic particles. However, this is a complex picture for novice learners to absorb.

Caloric Theory as a bridging analogy

David Hammer (2000) argues persuasively that certain common student cognitive resources can serve as anchoring conceptions because they align well with physicists’ understanding of a particular topic. An anchoring conception helps to activate useful cognitive resources and a bridging analogy serves as a conduit to help students apply these resources in what is, initially, an unfamiliar situation.

The anchoring conception in this case is students’ understanding of the behaviour of liquids. The useful cognitive resources that are activated when this is brought into play include:

  • the idea of spontaneous flow e.g. water flows downhill;
  • the idea of measurement e.g. we can measure the volume of liquid in a container; and
  • the idea of conservation of volume e.g. if we pour water from a jug into an empty cup then the total volume remains constant.

The bridging analogy which serves as a channel for students to apply these cognitive resources in the context of understanding energy transfers is the idea of ‘energy as a quasi-material substance’ (which can be considered as an iteration of the ‘adapted’ Caloric Theory which includes the conservation of energy).

The bridging analogy helps students understand that:

  • energy can flow spontaneously e.g. from hot to cold;
  • energy can be measured and quantified e.g. we can measure how much energy has been transferred into a thermal energy store; and
  • energy does not appear or disappear: the total amount of energy in a closed system is constant.

Of course, a bridging analogy is not the last word but only the first step along the journey to a more complete understanding of the physics involved in energy transfers. However, I believe the ‘energy as a quasi-material substance’ analogy is very helpful in giving students a ‘sense of mechanism’ in their first encounters with this topic.

Teachers are, of course, free not to use this or other bridging analogies, but I hope that this post has persuaded even my more reluctant colleagues that they need a more substantive argument than a knee jerk ‘energy-as-substance = Caloric Theory = BAD’.


References

Fairhurst P. (2021), Best Evidence in Science Teaching: Teaching Energy. https://www.stem.org.uk/sites/default/files/pages/downloads/BEST_Article_Teaching%20energy.pdfhttps://www.stem.org.uk/sites/default/files/pages/downloads/BEST_Article_Teaching%20energy.pdf [Accessed April 2022]

Hammer, D. (2000). Student resources for learning introductory physicsAmerican Journal of Physics68(S1), S52-S59.

Institute of Physics (2016), Physics Narrative: Shifting Energy Between Stores. Available from https://spark.iop.org/collections/shifting-energy-between-stores-physics-narrative [Accessed April 2022]

‘Killer-joules’ and other abominations

As noted earlier, some students struggle with unit conversions. To take a simple example: if we need to convert 3.7 kilojoules (or ‘killer-joules’ as some insist on calling them *shudders*) into joules, then whilst many students know that the conversion involves applying a factor of one thousand, they do not know whether to multiply 3.7 by a thousand or divide 3.7 by a thousand.

Michael Porter shared a brilliant suggestion for helping students over this hurdle. He suggests that we break down the operation into two parts:

  • Consider if we are making the unit larger or smaller.
  • If making the unit larger, we must make the number smaller to compensate; and vice versa.

Let’s look at using the Porter system for the example shown above.

(Note: I have used kilojoules for our first example since, at least for GCSE Science calculation contexts, students are unlikely to have to convert kilograms into grams. This is because, of course, the kilogram (not the gram) is the base unit of mass in the SI System.)

By changing from kilojoules to joules we are making the unit smaller, since one kilojoule is larger than one joule.

To keep the measured quantity of energy the same magnitude, we must therefore make the number part of the measurement bigger to compensate for the reduction in size of the unit.

This leads us to the final answer.

Now let’s look if we had to convert 830 microamps into amps:

The strange case of time

Obviously 1 minute is a very small quantity of time compared with a whole week. Indeed, our forefathers considered it small as compared with an hour, and called it “one minùte,” meaning a minute fraction — namely one sixtieth — of an hour. When they came to require still smaller subdivisions of time, they divided each minute into 60 still smaller parts, which, in Queen Elizabeth’s days, they called “second minùtes” (i.e., small quantities of the second order of minuteness).

Silvanus P. Thompson, “Calculus Made Easy” (1914)

It is probable that the division of units of time into sixtieths dates back many thousands of years to the ancient Babylonians(!) Is it any wonder that some students find it hard to convert units of time?

We can use the Porter system to help students with these conversions. For example, what is 7 hours in seconds?

This type of diagram is, I think, very useful for showing students explicitly what we are doing.

Units, you nit!

The S.I. System of Units is a thing of beauty: a lean, sinewy and utilitarian beauty that is the work of many committees, true; but in spite of that common saw about ‘a camel being a horse designed by a committee’, the S.I. System is truly a thing of rigorous beauty nonetheless.

Even the pedestrian Wikipedia entry on the 2019 Redefinition of the S.I. System reads like a lost episode from Homer’s Odyssey. As Odysseus tied himself to the mast of his ship to avoid the irresistible lure of the Sirens, so in 2019 the S.I, System tied itself to the values of a select number of universal physical constants to remove the last vestiges of merely human artifacts such as the now obsolete International Prototype Kilogram.

Meet the new (2019) SI, NOT the same as the old SI

However, the austere beauty of the S.I. System is not always recognised by our students at GCSE or A-level. ‘Units, you nit!!!’ is a comment that physics teachers have scrawled on student work from time immemorial with varying degrees of disbelief, rage or despair at errors of omission (e.g. not including the unit with a final answer); errors of imprecision (e.g. writing ‘j’ instead of ‘J’ for ‘joule — unforgivable!); or errors of commission (e.g. changing kilograms into grams when the kilogram is the base unit, not the gram — barbarous!).

The saddest occasion for writing ‘Units, you nit!’ at least in my opinion, is when a student has incorrectly converted a prefix: for example, changing millijoules into joules by multiplying by one thousand rather than dividing by one thousand so that a student writes that 5.6 mJ = 5600 J.

This odd little issue can affect students from across the attainment range, so I have developed a procedure to deal with it which is loosely based on the Singapore Bar Model.

A procedure for illustrating S.I. unit conversions

One millijoule is a teeny tiny amount of energy, so when we convert it joules it is only a small portion of one whole joule. So to convert mJ to J we divide by 1000.

One joule is a much larger quantity of energy than one millijoule, so when we convert joules to millijoules we multiply by one thousand because we need one thousand millijoules for each single joule.

In time, and if needed, you can move to a simplified version to remind students.

A simplified procedure for converting units

Strangely, one of the unit conversions that some students find most difficult in the context of calculations is time: for example, hours into seconds. A diagram similar to the one below can help students over this ‘hump’.

Helping students with time conversions

These diagrams may seem trivial, but we must beware of ‘the Curse of Knowledge’: just because we find these conversions easy (and, to be fair, so do many students) that does not mean that all students find them so.

The conversions that students may be asked to do from memory are listed below (in the context of amperes).

A table showing all the SI prefixes that GCSE students need to know

Dual Coding and Equations of Motion for GCSE

Necessity may well be the mother of invention, but teacher desperation is often the mother of new pedagogy.

For an unconscionably long time, I think that I failed to adequately help students understand the so called ‘equations of motion’ (the mathematical descriptions of uniformly accelerated motion using the standard v, u, a, s and t notation) because I suffered from the ‘curse of knowledge’: I did not find the topic hard, so I naturally assumed that students wouldn’t either. This, sadly, proved not to be the case, even when the equation was printed on the equation sheet!

What follows is a summary of a dual coding technique that I have found really helpful in helping students become confident with problems involving the ‘equations of motion’. This is especially true at GCSE, where students encounter formulas such as

Equation of motion (v^2 - u^2 = 2as)

for the first time.

A dual coding convention for representing motion

Table to show dual coding convention

Applying dual coding to an equation of motion problem

EXAMPLE: A car is travelling at 6.0 m/s. As the car passes a lamp-post it accelerates up to a velocity of 14.2 m/s over a distance of 250 m. Calculate a) the acceleration; and b) the time taken for this change.

The problem can be represented using the dual coding convention as shown below.

Diagram showing car accelerating past a lamp-post using the dual coding convention outlined in the blog

Note that the arrow for v is longer than the arrow for u since the car has a positive acceleration; that is to say, the car in this example is speeding up. Also, the convention has a different style of arrow for acceleration, emphasising that it is an entirely different type of quantity from velocity.

We can now answer part (a) using the FIFA calculation system.

Working showing solution for part (a)

Part (b) can be answered as shown below.

Working for part b of the problem

Visualising Stopping Distance Questions

These can be challenging for many students, as we often seem to grabbing numbers and manipulating them without rhyme or reason. Dual coding helps make our thinking explicit,

EXAMPLE: A driver has a reaction time of 0.7 seconds. The maximum deceleration of the car is 2.6 metres per second squared. Calculate the overall stopping distance when the car is travelling at 13 m/s.

We need to calculate both the thinking distance and the braking distance to solve this problem

The acceleration of the car is zero during the driver’s reaction time, since the brakes have not been applied yet.

We visualise the first part of the problem like this.

Using s=vt we find that the thinking distance is 9.1 m.

Now let’s look at the second part of the question.

There are three things to note:

  • Since the car comes to a complete stop, the final velocity v is zero.
  • The acceleration is negative (shown as a backward pointing arrow) since we are talking about a deceleration: in other words, the velocity gradually decreases in size from the initial velocity value of u as the car traverses the distance s.
  • Since the second part of the question does not involve any consideration of time we have omitted the ‘clock’ symbols for the second part of the journey.

We can now apply FIFA:

Working for braking distance part of problem

Since the acceleration arrow points in the opposite direction to the positive arrow, we enter it as a negative value. When we get to the third line of the Fine Tune stage, we see that a negative 169 divided by negative 4.4 gives positive 38.408 — in other words, the dual coding convention does the hard work of assigning positive and negative values!

And finally, we can see that the overall stopping distance is 9.1 + 38.4 = 47.5 metres,

Conclusion

I have found this form of dual coding extremely useful in helping students understand the easily-missed subtleties of motion problems, and hope other science teachers will give it a try.


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Signposting Fleming’s Left Hand and Right Hand Rules

Students undoubtedly find electromagnetism tricky, especially at GCSE.

I have found it helpful to start with the F = BIl formula.

Introducing the “magnetic Bill” formula

This means that we can use the streamlined F-B-I mnemonic — developed by no less a personage than Robert J. Van De Graaff (1901-1967) of Van De Graaff generator fame — instead of the cumbersome “First finger = Field, seCond finger = Current, thuMb = Motion” convention.

Students find the beginning steps of applying Fleming’s Left Hand Rule (FLHR) quite hard to apply, so I print out little 3D “signposts” to help them. You can download file by clicking the link below.

You can see how they are used in the photo below.

Photograph to show how to use the FLHR signpost

Important tip: it can be really helpful if students label the arrows on both sides, such as in the example shown below. (The required precision in double sided printing defeated me!)

Another example of the FLHR signpost in use

Signposts for Fleming’s Right Hand Rule are included on the template.


Other posts I have written on magnetism include:

Feel free to dip into these 🙂