n.b. This is "for interest”.
Suppose in the 5-mass case we need to find both angles: we simply know that they are A, -B, 0, B, -A. The two equations to be solved are
1 + 2 cosB + 2 cosA = 0
2(sinB - 2sinA) = 0
The other two equations (remember we have 4 - the two force components and two moment components) are automatically satisfied because of the symmetry of the A, -B, 0, B, A configuration.
We could solve this algebraically:
- square both equations
- substitute one in the other
- this gives a quadratic for one angle
- substitute to find the other angle
There are easier ways, though, if you as an engineer just want a quick solution. One is to use an iterative method:
beta = 360/5; % 72 degrees, initial guess
for i = 1:5,
alpha = 180 - asind(sind(beta)/2);
beta = acosd(-0.5-cosd(alpha));
disp([alpha beta]),
end
151.6062 67.6849
152.4479 67.2575
152.5402 67.2112
152.5503 67.2062
152.5514 67.2056
Try doing this on your calculator. Check it's in degrees mode, then:
72 =
180 - sin-1(sin(ANS) / 2) =
cos-1(-0.5 - cos(ANS)) =
Press “=" a few times, using the UP arrow each time to go back to the previous equation so you alternate between the “ 180 - …" and "cos-1(…” lines - watch it converge, giving you better and better estimates for the A and B values.
We could also use an algebraic solver such as the symbolic toolbox in Matlab. If you have the toolbox installed (type “help symbolic" to find out) you can simply type:
>> syms A B
>> eqn1 = 1 + 2*cos(B) + 2*cos(A) ==0; % note “=="
>> eqn2 = sin(B) - 2*sin(A) ==0;
>> AB = solve(eqn1, eqn2)
AB =
A: [4x1 sym]
B: [4x1 sym]
% AB is a “structure” with a field for each coefficient.
>> AB.A
ans =
-2*atan((11*(- (2*10^(1/2))/3 - 5/3)^(1/2))/2 + (3*(- (2*10^(1/2))/3 - 5/3)^(3/2))/2)
2*atan((11*(- (2*10^(1/2))/3 - 5/3)^(1/2))/2 + (3*(- (2*10^(1/2))/3 - 5/3)^(3/2))/2)
-2*atan((11*((2*10^(1/2))/3 - 5/3)^(1/2))/2 + (3*((2*10^(1/2))/3 - 5/3)^(3/2))/2)
2*atan((11*((2*10^(1/2))/3 - 5/3)^(1/2))/2 + (3*((2*10^(1/2))/3 - 5/3)^(3/2))/2)
>> eval(AB.A)
ans =
0.0000 + 0.6528i
0.0000 - 0.6528i
-2.6625 + 0.0000i
2.6625 + 0.0000i
% There are 4 solutions. We aren’t interested in the imaginary ones. The two real terms are the radian equivalents of the +/-152.6 degrees (A) and +/-67.2 degrees (B).
>> eval(AB.B)
ans =
3.1416 - 1.1382i
-3.1416 + 1.1382i
-1.1730 + 0.0000i
1.1730 + 0.0000i
Footnote:
Back in 1884 this kind of calculation was the basis for the Yarrow-Schlick-Tweedy system for engine balancing (we will see the relevance of shaft balancing to engines with pistons later, in balancing_app). Triple expansion marine steam engines could not have moment balance (only 3 cylinders): the solution was to convert the large LP (‘low pressure”cylider into two smaller ones, positioned at each end of the engine. The crankshaft for this 4 cylinder engine was rather like the above 5-mass shaft, with the central mass removed.
Nowadays it is mostly of academic interest. 5 cylinder car engines have 72 degree angles because this gives an even firing interval (the engine feels “rough” is the interval is uneven - this is acceptable on a motorcycle but less so for a luxury car).
Large marine engines (up to 18 cylinders) have crank angles carefully chosen to achieve both inertial balance and also to avoid exciting any torsional vibration modes.