Ninigishzidda !!! You are absolutely correct
I researched your line of thought and found this ....
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Imagine the object to be a spinning bicycle wheel, held at both ends of its axle in the hands of a subject. The wheel is spinning clock-wise as seen from a viewer to the subject’s right. Clock positions on the wheel are given relative to this viewer. As the wheel spins, the molecules comprising it are travelling vertically downward the instant they pass the 3-o'clock position, horizontally to the left the instant they pass 6 o'clock, vertically upward at 9 o'clock, and horizontally to the right at 12 o'clock. Between these positions, each molecule travels components of these directions, which should be kept in mind as you read ahead. The viewer then applies a force to the wheel at the 3-o'clock position in a direction away from himself. The molecules at the 3-o'clock position are not being forced to change their direction when this happens; they still travel vertically downward. Actually, the force attempts to displace them some amount horizontally at that moment, but the ostensible component of that motion, attributed to the horizontal force, never occurs, as it would if the wheel was not spinning. Therefore, neither the horizontal nor downward components of travel are affected by the horizontally-applied force. The horizontal component started at zero and remains at zero, and the downward component is at its maximum and remains at maximum. The same holds true for the molecules located at 9 o'clock; they still travel vertically upward and not at all horizontally, thus are unaffected by the force that was applied. However, molecules at 6 and 12 o'clock are being forced to change direction. At 6 o'clock, molecules are forced to veer toward the viewer. At the same time, molecules that are passing 12 o'clock are being forced to veer away from the viewer. The inertia of those molecules resists this change in direction. The result is that they apply an equal and opposite reactive force in response. At 6 o'clock, molecules exert a push directly away from the viewer, while molecules at 12 o'clock push directly toward the viewer. This all happens instantaneously as the force is applied at 3 o'clock. Since no physical force was actually applied at 6 or 12 o’clock, there is nothing to oppose these reactive forces; therefore, the reaction is free to take place. This makes the wheel as a whole tilt toward the viewer. Thus, when the force was applied at 3 o'clock, the wheel behaved as if that force was applied at 6 o'clock, which is 90 degrees ahead in the direction of rotation. This principle is demonstrated in helicopters. Helicopter controls are rigged so that inputs to them are transmitted to the rotor blades at points 90 degrees prior to the point where the change in aircraft attitude is desired.
Precession causes another phenomenon for spinning objects such as the bicycle wheel in this scenario. If the subject holding the wheel removes one hand from the end of the axle, the wheel will not topple over, but will remain upright, supported at just the other end of its axle. However, it will immediately take on an additional motion; it will begin to rotate about a vertical axis, pivoting at the point of support as it continues spinning. If the wheel was not spinning, it would topple over and fall when one hand is removed. The ostensible action of the wheel beginning to topple over is equivalent to applying a force to it at 12 o'clock in the direction of the unsupported side (or a force at 6 o’clock toward the supported side). When the wheel is spinning, the sudden lack of support at one end of its axle is equivalent to this same force. So, instead of toppling over, the wheel behaves as if a continuous force is being applied to it at 3 or 9 o’clock, depending on the direction of spin and which hand was removed. This causes the wheel to begin pivoting at the point of support while remaining upright. It should be noted that although it pivots at the point of support, it does so only because of the fact that it is supported there; the actual axis of precessional rotation is located vertically through the wheel, passing through its center of mass. Also, this explanation does not account for the effect of variation in the speed of the spinning object; it only describes how the spin axis behaves due to precession. More correctly, the object behaves according to the balance of all forces based on the magnitude of the applied force, mass and rotational speed of the object.
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It was somewhere between 9 o'clock and 12 o'clock when the snakling happened so it all adds up !!!
On a more serious note, reading about precession has made me realise that the issue
is probably in the rear wheel and not the front. When I bought the bike second hand, the honest seller warned me that the rear wheel 'needed truing' as he put it. I later realised that the frame has a fault in it which causes the rear wheel to pull to the left (when viewed from the rear) if it is merely tightened in the rear drop outs. I have had to hand make a tiny wedge out of aluminium that I insert between the hub and the drop out to kick the rear wheel closer to the centre line. However, I know that while this has stopped the tyre from rubbing the chain stay, the wheel is still not 100% where it should be.
At speed (I was nudging 40 mph down hill), this is probably creating a wobble that results in the frightening experience I had.
Any advice on how to better centre my rear wheel graciously accepted.
