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A Static Explanation of the Orthospinology C/A Factor
Roderic E. Rochester, D.C.
The C/A Factor is one of four factors that orthospinologists consider in the determination of a Height Factor. The four factors are: Plane Line; C/A; Atlas/Odontoid; and Angles Considering these four factors, plus the atlas rotation, a specific adjusting vector can be determined. Many questions are asked from students regarding the purpose of the C/A factor. Many times they have questions that are difficult to answer! One such question is: Why add 1/2 inch for every inch different the A circle is larger than the C circle? Why not add 1 inch, 3/4 inch etc.?
The only answer I have ever heard is that this is what Dr. John F. Grostic said to add! Unfortunately, this answer is insufficient in today's health care climate. I have asked many chiropractors using this procedure for the mathematical reasoning for this and all Height Factors. One doctor told me that some Georgia Tech engineers concluded there was no mathematical reason. Another told me it was developed empirically (meaning trial and error). Disbelieving this I set out to discover the reason behind the C/A Factor. The purpose of the C/A Factor is to determine the vectored force necessary, to apply to the C1 transverse process, that will transfer force to condylar and axial surfaces from the superior and inferior C1 surfaces, resulting in an equal rotation of the Central Skull Line and the Lower Angle.
Once this is visualized, it is easy to see that adjusting steeper than the C/A vector would create a lever situation with the larger force transferred to the axial surface. Vice versa, adjusting below or flatter than the C/A vector would create a lever with the larger force transferred into the condylar surface. For that reason I think of the C/A Factor as the "Great Equalizer!" Figure A supplies graphic representation for the calculation of that vector. The derivation for the C/A Formula is as follows:
Derivation of the C/A Formula
Angle Beta = Beta
Angle F = F
Angle D = D
Angle E = E
AngleY =Y
Angle Z = Z
Beta = F - D
F = .5 (D + E)
D = 90 -Y
Y = Sin-l (1/2 I C/2) = Sin-l (1/C)
E = 90 - Z
Z = Sin -1 (1/2 IA12) = Sin -1 (1/A)
so: Beta = [.5[(90 - Sin-l (1/C)) + (90 -Sin-l(l/A))]] - [90 - Sin-l(l/C)]
Using this formula and various C's and A's we find:
C/A Vector
3/4 2.49 degrees
3/5 3.97 degrees
3/6 4.94 degrees
3/7 5.63 degrees
3/8 6.15 degrees
3/9 6.55 degrees
3/10 6.87 degrees
3/11 7.13 degrees
3/12 7.35 degrees
3/8 or 5.68 degrees Average, so:
8A - 3C is 5 thus 5.68 / 5 = 1.14 degrees for every inch A is larger than C
Convert 1.14 degrees to inches of height factor
1.14 deg. =Sin (X / 24)
Sin-1 (1.14) = X / 24
Sin-1 (1.14) * 24 = X
X = .48 inches of Height Factor on average for every inch difference A is larger than C. This is where the 1/2 inches comes from!
But is this formula correct? After all, there are other theories in similar techniques that have a different C/A Factor formula.
To answer this question I built a model of the skull, atlas and lower angle from wood. I built this model to scale with a 3 C / 8 A set up, but large enough to clearly see and measure changes that occurred from applying adjustive forces at various lines of drives. I stabilized the C7 point of the wooden lower angle with a screw, because for all practical purposes this point is "fixed" in real adjusting situations. I stabilized the mastoid with another screw to simulate the mastoid support. With the model in a neutral position I applied a force to the transverse process of C1 with our hand-held adjusting instrument (power turned up). The adjustive force was applied at angles varying from - 24 degrees to + 90 degrees. At each angle, three "adjustments" were performed, returning the model to neutral before each thrust. Each thrust created a "misalignment" of the Central Skull Line and the Lower Angle. I carefully measured and recorded the results of each thrust at each angle. I averaged three trials for each line of drive. The results were exciting.
I plotted on a graph, the changes that occurred to the lower angle and central skull line with the varying lines of drive. Figure B shows the graph.
Three important things can be learned from this chart:
1. If the formula is close, for the C/A vector, the Lower Angle Line and the Central Skull Line, the line should cross at the calculated vector angle for a 3C/8A which is 6.15 degrees. The point at which the lines cross from the data is 5.6 degrees, a 9% difference. In other words at a 5.6 degree vectored thrust, an equal change of 1.4 degrees, occurred for the Central Skull Line and the Lower Angle Line in the model. This demonstrates that the C/A Factor, as we calculate it, provides the best line of drive for equal reduction of the Central Skull Line and the Lower Angle Line!
2. Maximum leverage occurred with a vector angle of 36 degrees, however, there was very little difference between the 24, 30 and 36 degrees thrust.
3. From the 36 to the 48 degree angle vectors, leverage is significantly REDUCED. After 48 degrees, leverage is sharply reduced and at the 90 degree thrust no significant change in alignment could be recorded.
This exercise is not meant to be the final word, but a starting point. Hopefully, the mechanical engineers and mathematicians can develop or refine a formula. It is interesting to note that in the chart below, the graph lines crossed at a point 9% below where the math formula predicted, but they crossed exactly where J. F. Grostic, D.C. said they would!
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