I wanted to arrange the LEDS on my machine equally spaced around a circle, as shown in the image above.  I couldn’t use a protractor as the inside of the enclosure is too small, but I could use drawing compasses.  I decided that I needed twelve points around the circumference of a 100 mm diameter circle. But how to work out where to place the marks?

Firstly draw the circle – 100 mm diameter means set the compasses to a radius of 50 mm.  Now there are 360 degrees in a circle, so there will be 360/12 = 30 degrees between each LED.  How can I get this?

If you wrack your brains you may remember the Sine rule from your school geometry.  I’ve written it out below and I’ve shown it on the diagram.

a/sin(A) = b/sin(B) = c/sin(C)

where a, b and c are the lengths of the three sides of the triangle and A, B and C are the angles opposite these sides.

From this, the length of the side a that we want is given by

a = b*sin(A)/sin(B)

The sine is a mathematical function that relates the ratio of the lengths the sides of a triangle to the angles enclosed. It has many uses, originally in surveying, and has sister functions Cosine and Tangent.  We don’t need to go  there now, only to know that’s why these ‘trigonometric functions’ appear on all scientific calculators.

Angles round circle

Since the angles inside a triangle add up to 180 degrees, and the angle A that we want is 30 degrees, the two angles opposite must add up to 150 degrees. Since the angles are both the same, they must each be  B=C=75 degrees.  And we know that the two equal sides b and c are both 50 mm long (the radius of the circle).  Using a ‘scientific’ calculator that has sines on it, we can work out that the length of the opposite side is 25.88 mm.

Set your compasses to this value and mark off this distance around the circle, starting from a suitable diameter mark.  The sixth mark should exactly sit on the opposite end of the diameter.  If it is only a tiny bit out, mark backwards from this end, back towards the beginning.  The exact positions with be exactly between the marks made from opposite ends.  If you’re not happy, adjust the compasses until these exactly coincide from both ends.

You can use the same method to mark any angle you wish – for example ten points round a circle will be at 36 degree spacing. You’ll need to recalculate the angles b and c, each of which will be
(180-36)/2 = 72 degrees.

There are plenty of circuits showing how to control a single LED from the Pi, but Sirima needs to switch between a red and a green LED and I only want to use one pin.

After some thought, I am using the circuit below

The left hand side of the circuit is the usual circuit that uses an NPN transistor to turn the left-hand (red) LED on when the GPIO pin voltage goes high (to + 3v3).

It seems obvious that you could use a PNP transistor to turn a different LED on when the pin goes low (to 0 v)but this doesn’t work well because there is always a ground path through the pin via its built-in protection diodes, and in any case it uses an extra transistor.

However, there is another way to deal with this, as shown in the right-hand side of the circuit.  This LED is connected to the current-limiting resistor (i.e. to the same point as the anode of the left-hand LED).  It uses a 2.4 v zener diode to limit the voltage across the right-hand LED (which requires 2.6 volts for full brightness).   When the transistor switches on, it parallels the ground path through the right-hand LED so this turns off.  Without the zener, the current would be shared between the two LEDs and they would both illuminate weakly.  Ideally the zener voltage should be a bit less than 2.4 v but I couldn’t find one readily.  [An alternative would be two use ordinary silicon two diodes in series in the conducting direction (anode towards the positive), as they each have a drop of about 0.7 v, giving 1.4v in total]  As it is, the current through the green LED is less than rated, but this doesn’t matter in my application, as the green LED is brighter than the red one at their rated current.

Not only does this circuit minimise the components needed, but only three wires are needed to operate the two diodes.

It should also be noted that in the Raspberry Pi, the state of the GPIO pins is undefined during boot-up and so it is rather random as to which LED will be illuminated at this time. To switch them all off would need a MOSFET, say, to control the 5v rail, needing another pin to be used.

My ringing machine – I’m thinking of calling it Sirima (Silent Ringing Machine) – is ready for prototyping. I’ve bought a plastic case (so the Pi’s built-in WiFi and Bluetooth can still work).  The box will house a loudspeaker, so it will only be monaural, but I don’t need more for this prototype. It will also use a circle of LEDs to show the bells as they sound. The illustration shows it part-made.  I will need to use transistors as switches to turn the LEDs on and off as the Pi’s GPIO pins can’t supply enough power.

Originally I thought there would be  eight bells, I’ve made space for 12 bells.  There will be a green LED for handstroke and a red one for backstroke,  arranged in two concentric circles, but the prototype will use just eight pairs.

The back panel of the box will house 3-pin DIN connectors for the photocell sensors, which need a 5 v power supply and have one ‘open collector’ output wire.  I have five sensors, but will just connect one to start with, as there are quite a few practical issues with mounting the sensor and routeing the cable.

I’m using a 12-watt audio amplifier module with a 4-inch (100 mm) loudspeaker.  This needs a 12-volt DC smoothed power supply and I have bought a small switched-mode 2-amp power supply.  I have a small potentiometer that I can use as a volume control which will need to be on the front panel.

I’d like a single mains power inlet for both the Pi and the amplifier.  A Pi model B (which I’m using in the prototype) can take between 0.7 to 2.5 A at 5V.   The board typically needs 0.5A.  The normal way of supplying power is via the micro-USB plug , but you can supply the power via Pin 2 (+5 V) and Pin 6 (Gnd), although there is no regulation or current protection on these.  I will also need power for the photosensors and LEDs.  From experience, it might be better if these are not powered directly from the PI’s GPIO pins as they shouldn’t need a lot of additional smoothing as long as there are no spikes on the power supply.  So I’m thinking of getting a 5v switched power supply for all this. I can then fit a mains inlet and power switch onto the back panel.

In its simplest mode, the system can just emit a bell sound whenever the motion sensor is triggered.  However, in order to practice bell control, it is necessary to have, say, six bells but with a gap for the student’s bell to ring in.  This would need some sort of start/stop button if it is running headless, or else it would be necessary to rdp into it from a laptop to do the setup.  Of course, it would be nice to have other features such as a speed control switch, but this is some way down the road.

Anyway, I’m taking a break for the next few days, so more in a little while.