Vehicle Dynamics & Control - 09 Dynamic bicycle model with linear tires

[@Waka-bm4ce]

Thank you for these courses, really well explained

[@readmeCarefully]

Very informative and clear explanation

[@sureshnayakbhukya5911]

Sir, can you tell how to derive equations of motion for Tractor-Trailer vehicle during steering?

[@josephlai7737]

5:46 How is a_y(X,Y) derived? Why are there two terms? Is it because the car's motion a combination of translation and rotation, and the two terms correspond to translational and rotational acceleration respectively? I though centripedal acceleration is omega * v, and the direction is pointing to the instantaneous center of rotation, which does not align with vehicle lateral direciton. Or is it because the second term is the result of omega * v * cos(beta) = omega * v_lon (takign the vehicle lateral component of the total centripedal acceleration)?

[@eaglezhou1243]

Excellent!

[@SoumilSahu]

Hi, great video! I have a question though: The assumption that v_lon is a constant immediately seems to break down if we write the equation of motion in the x (lowercase) direction. There is a force, ie; F_f * delta , which is unaccounted for. What am I missing here?

[@professorschildbach]

You are not missing anything. We are using the model only for lateral dynamics of the vehicle. Like any model, the DBM is not fully correct, also not for the lateral dynamics. For example, it assumes that the car has two tires and no rolling motion, which is obviously wrong. Here we do not care about the model in the longitudinal directions. Therefore we do not resolve this (apparent) contradiction and leave it open. If you care about it, you would need to assume, for example, an additional (accelerating) force in the longitudinal direction, e.g., via the rear tire. You can assume this longitudinal force compensates exactly the braking component of the steering force, if that makes you happy.

[@elmehdinaor763]

Prof.Georg Schildbach , Sie haben einen sehr guten Job gemacht. Viele Grüße aus Kaiserslautern :D

[@professorschildbach]

Thank you for your comment, really appreciated. Hope the material is useful! Suggestions are very welcome.

[@kataimwanza5295]

Fantastic teaching Professor

[@utkukurt9601]

Allah razi olsun

[@kingsgambit]

A question regarding the rotational inertia I_z: We would calculate that with I_z = m * R^2 (since we model the vehicle as a point mass), with R given by the expression for R from the kinematic model, is that correct?

[@professorschildbach]

No. A point mass has I_z=0. You must take the value of I_z that is given for the vehicle. The formula "m*r^2" would be correct for single mass points, but you need to sum up over the entire vehicle. In reality, I_z must be determined experimentally or estimated.

[@kingsgambit]

@@professorschildbach A point mass has I=0 ??? I learned that a point mass' inertia depends on the position relative to the axis of rotation. Maybe I misunderstood you. I understand that the way you describe is correct to get the "real" rotational inertia, but can it not be modelled in a simplified way as a point? If not, why?

[@professorschildbach]

@@kingsgambit I_z always refers to the CoG of the system. If not, it would depend on the rotational axis, and hence not be a pure property of the system itself.

[@kingsgambit]

@@professorschildbach thank you!

[@pilotar6701]

Hello Professor Schildbach, I am confused by the coordinate transformation done in this video. Seems to me to go from the xy direction to the e,n direction, the rotation matrix should be set up as [cosd, -sind; sind, cosd], which gives V_ae= V_ax cosd -V_ay sind, and V_an=V_ax sind+Vay cosd. What am I doing incorrectly here? Edit: Did my matrix multiplication wrong. It has been corrected.

[@10gala10]

See en.wikipedia.org/wiki/Rotation_of_axes

[@pilotar6701]

@@10gala10 thanks for the reply. I had the Y - Axis flipped from what is the standard SAE convention, so all of my signs were different and not matching what was done in the video.

[@10gala10]

@@pilotar6701 For me, the important point is to differentiate a rotation matrix from an axis rotation. The former translates the final position of a point (or vector), after a rotation, in the original coordinate system. The latter rotates the coordinate system, thus, translates a point (or vector) from the original coordinate system to the rotated one. The direction of your axes is a different subject, but it can indeed affect your results and make them non-equivalent to the video's result.

[@pilotar6701]

Hi @@10gala10. A rotation matrix rotates a vector about the origin of the coordinate system around an axis to get the new coordinate system, there is no translation as you described above. That is what is going on in the Wikipedia link you sent me. They derive it two ways, using trigonometry, or using matrix math. What you are describing as a rotation matrix is a transformation matrix that includes rotation column vectors, and a translation column vector. That being said, a transformation matrix can be set up in such a way that there is only pure rotation, pure translation, or both. In the case of both, you correctly pointed out that the rotation occurs first, and then the coordinate system is translated. As for the incorrect axis direction, once I flipped the Y-Axis and changed the rotation matrix accordingly, I obtained the same result as the video.

[@10gala10]

Hi @@pilotar6701. Sorry, I used the word "translation" as converting from a language to another. Transforms, converts, express in terms, of would have been better choices, my bad.

[@user-jt2zr8wy5t]

Sir, I have a question. Why would you make an assumption of the small steering angle and why would sine and tanjent of it be delta, not 0?

[@professorschildbach]

Small steering angle assumption: We are engineers. It's a choice to make this assumption. You make in cases where you deem the assumption of a "small" steering angle as justified (e.g., on highways maybe?). Small angle approximation: These are the first order Taylor approximations of the sine and tangent functions.

[@irshadsmart98]

6:49 ,how do we know its clockwise or anti-clockwise to determine the sign

[@georgschildbach7118]

The relevant reference point is C, so you need to determine the direction of the rotation effect of a force with respect to C. If it is still not clear, please look for a basic mechanics book. I also have some videos on this ("Technische Mechanik 1"), but unfortunately they're in German.

[@lorenzonap123]

Sir , how do you derive the formula of centripetal acceleration 5:47 in this case? Pls help me

[@professorschildbach]

This is based on some formulas from basic mechanics: - angular speed omega = v / r - centripetal acceleration = v^2/r = omega^2*r

[@francescoindolfo]

Why there is a minus sign for the tire lateral forces ??

[@professorschildbach]

To see this, please imagine a positive value of the slip angle, and what the direction of the corresponding tire force will be. In fact, the tire force will point to the right, so in the opposite direction compared to the arrow drawn in the image. Conversely, for a negative slip angle, the tire force will point to the left, so in the same direction as arrow in the image. Hence the minus sign.

[@francescoindolfo]

@@professorschildbach ok so we have a positive slip angle when the y components of the wheel velocity is in the same sense as the positive y coordinate axe of the vehicle right??

[@professorschildbach]

@@francescoindolfo Yes, one could say so, except that due to the steering angle the wheel may not be aligned with the vehicle coordinates.

[@ismaelsantana7430]

Which biography are you using for these lectures? Thanks!

[@s.prudhvi7882]

4:45 is it vehicle frame or inertial frame sir?

[@michaelmccullough1646]

Also trying to understand that - seems like the forces are resolved into the vehicle frame? Unless im missing something

[@s.prudhvi7882]

@@michaelmccullough1646 yeah

[@erdemyusufgokduman4234]

In order to apply Newton's Second Law, all the forces and accelerations must be represented on an inertial frame. Since the vehicle frame is not an inertial frame, used frame must be global inertial frame(maybe fixed to the ground) .

[@narasimg]

How is psi calculated? It doesn't seem to depend on omega or v....

[@narasimg]

oh sorry psi (.) = omega