Rotated Reference Frames and Angles/Accelerations

[pgm575]

Homework Statement

Let's say that there is a bicycle traveling forward and we only see it from behind. As it rolls to turn, it induces some accelerations. If we were to measure those accelerations on the bicycle frame itself, we would see a Lateral Acceleration (lets call it Aym) and and a Vertical Acceleration (lets call it Azm). These two accelerations would be seen as having angles 0 deg and 90 deg (respectively) on the reference frame of the bicycle.

We want to know: 1) the value of the lateral acceleration on the world frame (or parallel to the ground). 2) the angle at which the bike is leaned over (or the angle from vertical).

What is given is Aym and Azm (the lateral and vertical accelerations on the bike-rotated coordinate frame); let's say for the sake of this discussion that they are 1G and 0.5G, respectively.

What is not given is the roll angle of the bicycle, or the lateral acceleration as measured against the world frame.

Homework Equations

For a generic world frame, with Theta being the angle from vertical, Theta = ArcTan(Lat Accel / Vert Accel).

Any triangle trig identities are also applicable, specifically Pythagorean Theorem.

The Attempt at a Solution

It's easy enough for me to draw out a resultant vector, which is 1.12G.
It's also easy enough to deduce the angles above and below the resultant vector within the bicycle-rotated coordinate frame, which are 26.6 deg and 63.4 deg, respectively.
I just don't get any farther.

 

[PeterO]

Homework Statement

Let's say that there is a bicycle traveling forward and we only see it from behind. As it rolls to turn, it induces some accelerations. If we were to measure those accelerations on the bicycle frame itself, we would see a Lateral Acceleration (lets call it Aym) and and a Vertical Acceleration (lets call it Azm). These two accelerations would be seen as having angles 0 deg and 90 deg (respectively) on the reference frame of the bicycle.

We want to know: 1) the value of the lateral acceleration on the world frame (or parallel to the ground). 2) the angle at which the bike is leaned over (or the angle from vertical).

What is given is Aym and Azm (the lateral and vertical accelerations on the bike-rotated coordinate frame); let's say for the sake of this discussion that they are 1G and 0.5G, respectively.

What is not given is the roll angle of the bicycle, or the lateral acceleration as measured against the world frame.

Homework Equations

For a generic world frame, with Theta being the angle from vertical, Theta = ArcTan(Lat Accel / Vert Accel).

Any triangle trig identities are also applicable, specifically Pythagorean Theorem.

The Attempt at a Solution

It's easy enough for me to draw out a resultant vector, which is 1.12G.
It's also easy enough to deduce the angles above and below the resultant vector within the bicycle-rotated coordinate frame, which are 26.6 deg and 63.4 deg, respectively.
I just don't get any farther.

I am having difficult recognising / accepting that there is any vertical acceleration?? The bike is getting no closer, and no further from the surface, so the is a constant 0 velocity - from a vertical point of view.?

 

[pgm575]

The vertical acceleration would be seen from the bike's reference frame. If there was only lateral force (on the world frame) when the bike is leaned over and it is accelerating laterally, transferring that force to the bike's frame of reference would introduce a vertical component. Thus the reason that the suspension of a bicycle or motorcycle compresses in cornering.

 

[pgm575]

Thinking about it another way maybe makes more sense- if your view was aligned with the bicycle's and you were rolling around a corner, you would seem to be traveling uphill (as your are traversing a curved path). Traveling uphill on a flat surface would also induce a vertical acceleration.