A diagram showing the arrangement is shown below. It is a rather complicated example, which will require some tricky mathematical description.
An electric motor has a gear wheel welded to the end, which has a radius r m. Note that when working with gear wheels, the effective radius is the point at which the teeth touch the teeth of the interlocking gear wheel. This will be somewhere between the radius to the end of the teeth, and the radius to the base of the teeth.
The motor’s gear wheel turns wheel 1, which has a radius of r1. This drives wheel 2 through the coupling rod of length ‘b’. Wheel 2 is also a gear wheel which drives a third gear wheel, wheel 3. The radii of wheels 2 and 3 are r2 and r3 respectively. The distance between centres of wheels 2 and 3 is ‘x’. The points of contact of the coupling rod on wheels 2 and 3 are ‘a’ and ‘c’ from the centre respectively.
The axe handle is attached to wheel 3, and is of length ‘q’. A mass of M is attached to the end of it. The mass of the axe handle may also be taken into account to arrive at a moment of inertia of the axe of Ia.
What we will be calculating
The system is designed so wheel 2 may rotate through an angle of 60° causing the axe to swing back and forth. We will work out the load torque on the motor required to perform this action, which will tell us the current required to drive the motor, and given a motor specification, the speed that the axe will swing at, and the energy it will have at the end of its swing.
Modelling the coupling rod
A complicated part of the modelling process is wheel 2 driving wheel 3. What we need to know is how much wheel 3 will turn for a given angle of turn of wheel 2. First we’ll sketch the arrangement as a line diagram with dimensions labelled. Angles are labelled with capital letters and lengths with small letters. Another line, d, has been drawn to split the quadrilateral shape into two triangles. Triangles are easier to deal with mathematically because we can make use of what is known as theCosine rule. Note that the name of an angle is the same as the name of the opposite edge of the triangles.
What we need to know is how much wheel 3 rotates when we rotate wheel 2. In terms of the wire diagram above, this translates to "how much does angle E change when angle D is changed?"
We can apply the cosine rule to the two triangles above as follows:
and knowing those we can find out E since:
The length d can be calculated again using the cosine rule as follows:
replacing d gives
which simplifies to
This looks rather complicated but remember all the dimensions a, b, c, and x, are known and so replacing these with number greatly simplifies the equation. This equation is modelled in the spreadsheet CouplingRod.xls (Excel 97 format).
In that spreadsheet, given the lengths a=1, b=10.05, c=2, and x=10, over the specified range of 60°, i.e. setting D from 60° to 120°, the angle of wheel 3 moves almost linearly as shown in the graph below:
This relationship can be modelled linearly for simplicity by the equation:
Remember that these numbers are specificly for the lengths entered in the spreadsheet.
Modelling the rest of the system
The rest of the system is much simpler than the coupling rod, since a change in angle of one wheel is linearly related to a change in angle of the next wheel. Going through the system from motor to axe, we can relate the angular movements as follows:
The second equation above seems to have lost its constant 37.348°. That’s because we’re now interested in how much the angle of wheel 3 changes when wheel 2 changes. The change in wheel 3 angle given a change in wheel 2 angle from 0° to 1° will be:
Therefore the constant is cancelled out, and that’s why it doesn’t appear in the equations above.
Merging all three equations above, we arrive at an equation which relates how much the axe moves when the motor shaft moves:
This is useful when we consider the load torque. This equation is effectively the gearing ratio for the system. If the axe is presenting a load torque of TL, then the torque at the motor shaft is:
As the axe swings around, it gains angular speed. There will be some effect of gravity since at the start of the stroke, according to the system diagram the axe head is moving uphill, and towards the end of the stroke it is moving downhill. The effect of gravity complicates the mathematics, and can safely be neglected since it will be relatively small if a powerful enough motor is used.
Moment of inertia of the axe
We are modelling the axe as a handle with a mass on the end of it. To simplify the calculation, we shall assume the mass is a point load at the end of the handle.
The equations for moment of inertia of a point mass, and of a handle are:
Where MP is the mass of the point load, MH is the mass of the handle, and q is the length of the handle.
Therefore the total moment of inertia of the axe is:
Speed, torque and angle
We can calculate the speed of the axe in a similar manner to the calculation for the spinning disk weapon presented here . Part of the way through that calculation, assuming a permanent magnet motor, we arrived at the equation:
where Ω1 is the no load speed of the motor, and T1 is the stall torque of the motor. We can rearrange this equation to give us the angular speed of the motor with respect to time:
For a motor with a no load speed of 3000 rpm and a stall torque of 10Nm, the speed increase with time is drawn as the red line in the graph below.
This shows the speed running up to near the no load speed. Torque is a linear function of speed in a PM motor,
The graph shows the torque reducing as the speed increases.
The instantaneous angle is calculated as follows. The angular speed is the rate of change of angle with time…
This integrates to
K is the constant of integration, and we will deal with that later.
The angle in the equation above is for the motor shaft. Remember from further up that the angle of the axe is derived from the angle of the motor shaft by the equation
Therefore we can merge these two equations to give us:
The const value here is not the same value as the K in the equation above, but is K multiplied by 0.4908(r2rm)/(r3 r1). All integration results have a have a constant added to them. Usually it is quite obvious what the value of this constant should be. In this case, before the axe starts to move, t=0, so placing that in the equation above gives us:
We know the starting angle of the axe – it is part of the mechanical design. All the other values in this equation are also known, so the constant can be calculated quite easily:
Replacing this into the large equation above yields:
Although this equation looks nightmarish, it is just a matter of plugging in the values from the mechanical design and motor characteristics, then plonking them in a spreadsheet. This is what I have done to create the graphs above.