Finding Derivatives for Bicycle Trajectory Plot in 2D

[twowheelsbg]

Homework Statement

Recently I met a Mathcad calculation plotting bicycle path traveled.
Initial positions of tire tracks are known, their increments were calculated with
bike speed 'v', wheelbase 'p', and steering angle 'a' so via ODE system solving
front and rear tracks coordinates were derived as time functions.
All ok, but I cannot comprehend how the derivatives are found.

Here is a simple sketch explaining letters in use:

Reference frame center is R0, coinciding with rear wheel contact point initially.
Bike is oriented in x-axis, front wheel contact at F0 initially at distance 'p' from R0,
steering is set right at angle 'a' assumed constant for simplicity.
I believe both wheels follow circular path with common center C
(maybe this is my problem )
R1 and F1 are the amended positions of R0 and F0 respectively after elementary time interval dt. Below also with 'b' is indicated angle between amended bike frame R1F1 and x-axis, obviously this angle increases gradually.

Homework Equations

as per the author of the Mathcad document:
dxr/dt={2.v.cosa/(1+cosa)}.cosb
dyr/dt={2.v.cosa/(1+cosa)}.sinb
dxf/dt={2.v/(1+cosa)}.cos(a+b)
dyf/dt={2.v/(1+cosa)}.sin(a+b)

The Attempt at a Solution

Bold members are reasonable - if tire track point increment is found,
it can be amended with the bold members to find their projections over the axles x and y,
rear having angle 'b' as bike frame is displaced from x-axis so, front angle is (a+b) as front wheel is misaligned from x-axis so.

Rear track radius is CR0=p/tga,
front track radius is CF0=p/sina , as they differ with cosa that means to me -
if i find one of the increments front F0F1 or rear R0R1, amending accordingly with cosa would give me the other. So I try to find the rear one as it seems easier to me, author idea about it as seen from above to be {2.v.cosa/(1+cosa)}.

As rear contact point travels via circle with tangent speed v, angular speed comes v/CR0,
elementary sector covered comes (v/CR0).dt ,
and elementary travel of rear point comes R0R1 = 2.CR0.sin{(v/CR0).dt/2} ... which simplified to first order ( sinx to x ) comes to v.dt ...
close to v.dt.{2.cosa/(1+cosa)} but not exactly

So, I am baffled ... pls help

 

[twowheelsbg]

I thought a bit, and changed my mind.
If my graphical interpretation is correct,
rear contact follows path with p/tga radius,
front contact follows path with p/sina radius,
and all intermediate points between R0 and F0 follow intermediate circles with intermediate radius, ... and intermediate speeds

So, if speed is given as 'v', this speed should count for masses center (M),
not for rear contact as I did before. Path radius CM0 comes from
CM0^2=CR0^2 + MR0^2, assume M is at middle, R0M=p/2,
so CM0=(p/tga).{1+(tga/2)^2}

Knowing path radius and speed for center of masses,
instant angular speed of rotation of M via C ( whole bike also ) is
w=speed/radius=(v.tga/p)/{1+(tga/2)^2}

And knowing this angular speed and front and rear radius, speeds are found:
Vr=w.p/tga=v/{1+(tga/2)^2}
vf=w.p/sina=(v/cosa)/{1+(tga/2)^2}=Vr/cosa

these can be projected to x and y axles easy,
observing that Vr is angled 'b' to x-axis, Vf is angled (a+b) to x-axis.

Obviously my result differs from expected : 1/{1+(tga/2)^2} is not equal to 2cosa/(1+cosa),
although difference is negligible for small steering angles.

Here I am so far ...

 

[twowheelsbg]

Negligible differences between my result and expected,
led me into doubt that something might be rounded for simplicity -
calculation of center of masses path radius as medium from front and rear radius
gave exactly the original results. Case closed