This post is the next in an installment of posts addressing the P4T flight path mathematics and begins with
this post. The first post defined centripetal acceleration as it is used by P4T in their calculation of bank angles and g-forces. This post expands and applies that concept into the vector components involved. I am using standard notation in this post, but it should be kept in mind that the notation applies to a vector and are differentials in one form or the other. Also, I am not familiar with vector notation in latex, so understand the notation is referring to a vector quantity.
The image above is a two-dimensional (x, z) for identifying the force vectors under discussion. It is important to understand that [latex]$$ F_L $$[/latex] (black arrow) represents lift, which is a function of [latex]$$ v_y $$[/latex] and [latex]$$ \sum F_y $$[/latex] which is not described here. As such, it is assumed (as in the P4T flight path) that [latex]$$ v_y $$[/latex] is constant and [latex]$$ \sum F_y = 0 $$[/latex].
Centripetal force ([latex]$$ F_c $$[/latex], red arrow) is the P4T favorite the past few months, so that is the place to start. [latex]$$ F_c $$[/latex] is not some magical force that comes out of nowhere and results in a circular flight path. Quite the contrary, it is the result of the flight path and the force has to come from somewhere. in this case, it is the [latex]$$ F_x $$[/latex] (blue arrow) component of [latex]$$ F_L $$[/latex]. Axis [latex]$$ z $$[/latex] is drawn parallel to the direction of the [latex]$$ F_c $$[/latex] vector, and [latex]$$ \theta $$[/latex] is the angle from [latex]$$ z $$[/latex] to the [latex]$$ F_L $$[/latex] vector.
[latex]$$ F_x = {F_L}sin{\theta} $$[/latex]
The component vector [latex]$$ F_x $$[/latex] must ALWAYS equal [latex]$$ F_c $$[/latex] or the path will not be circular. Also of interest is that the [latex]$$ F_z $$[/latex] (yellow arrow) component vector must ALWAYS be equal to and opposite of [latex]$$ F_g $$[/latex] (green arrow) if a horizontal path is to be maintained. This is the assumption of the P4T analysis which will become clear later.
[latex]$$ F_z = {F_L}cos{\theta} $$[/latex]
Since [latex]$$ F_z = m{a_z} $$[/latex] and [latex]$$ F_g = m{a_g} $$[/latex] and [latex]$$ m{a_z} = -m{a_g} $$[/latex], we can drop the mass quantity [latex]$$ m $$[/latex] and work directly with the acceleration vector.
Of course what P4T was wishing to determine was the bank angle [latex]$$ \theta $$[/latex]. The bank angle can be determined with the following equation based on the component vectors.
[latex]$$ tan{\theta} = \frac{a_x}{a_z} $$[/latex], or [latex]$$ {\theta} = {tan^{-1}}\left({\frac{a_x}{a_z}}\right) $$[/latex]
Since [latex]$$ a_x = a_c = \frac{v^2}{r} $$[/latex] and [latex]$$ a_z = -g $$[/latex],
[latex]$$ \theta = tan^{-1} \left({\frac {\left(\frac{v^2}{r}\right)}{-g}}}\right) $$[/latex]
Of course someone is going to scream, "that is not the equation they used!" I'm not going to do everything for P4T, so guys just plug in the conversion factors to make the equation specific to knots for [latex]$$ v $$[/latex] and feet for [latex]$$ r $$[/latex] see what you come up with. Also recall that g is in the negative direction, so the negative sign converts it to positive.
Now I went through all of this so that it becomes apparent to anyone paying attention that,
[latex] {F_L}sin{\theta} = m{\frac{v^2}{r}} $$[/latex] so,
[latex] {a_L}sin{\theta} = \frac{v^2}{r} $$[/latex]
In other words, the approach used by Mackey and I in rebuttal to P4T's previous venture regarding the official flight path being impossible was the correct approach. The isolation of the vertical component is a better estimate than the centripetal acceleration approach since neither [latex]$$ v $$[/latex] nor [latex]$$ r $$[/latex] could be considered constant. This is enough for this post, but next I will use components to do the same banking angle calculations with a diminishing altitude.
