I have never seen this degree of clarity in my life, not in kinematics at least. I did not want to comment again so as not to annoy, but I had to one more time. Once again Thank you so much.
There are multiple ways to derive the equations. All of them lead to the same result, of course. It is up to you, which way of calculation you find easier or more intuitive.
Thank you Prof Schildbach for this video. I have a question that we know the kinematic of a bicycle can be written as a nonlinear system like \dot x = f(x,u) and how to check controllability of this system?
Thank you Professor, very clear. I have a question with the last step in the derivation. Unfortunately my high school trig is lacking. So to anyone reading this, if you could enlighten me it would be most appreciated (exam's next Tuesday :/). How do you get from (1/lr+lf)*tan(d)*cos(B) ---to---> B = atan[(lr/lf+lr)*tan(d)] ?
@@professorschildbach Ah ok, so you just used the tan(B) which = lr/R(||) and then substituted in for R(||). The switch from cos to tan threw me for a second. But because the angle B can be expressed in any of the following ways sin(B)=lr/R, or cos(B)= R(||)/R, or tan(B)=lr/R(||) it's all the same. And you needed to find R first that's why used cos(B) instead of going straight to tan. Thanks for getting back to me
It is difficult to give a general answer. The the rear axle is typically used as a reference point for low-speed applications (like parking). The CoG is often used as a reference point for high-speed applications. The front wheel is rarely used, in my experience. However, it is used, for example, as the reference point for the Stanley Controller.
First, a normal car will not allow a steering angle of 90 degrees. That being said, the intuitive answer to that question should be clear: The vehicle will rotate around the center point of the rear wheel. Now, what happens mathematically is a little more interesting and requires to calculate the limit of R as delta -> 90 degrees. With the formulas in the video, it should give you l_r, but I have not tried.
@@professorschildbach TY for your response. I am using the model for an agv (one front wheel, steering not compareable to ackermann steering from a car). When I drive the agv with steering 90° it rotates around the rear wheel, just like you said. I am having trouble to model this mathematically. when delta -> 90° then tan(delta) -> inf => Therefore: omega (angular velocity) -> inf In my model i integrate omega to get the angle phi. And I use phi to calculate d/dt*x and d/dt*y (both should be zero, I take the rear wheel as origin of the coordinate system) -> as omega is inf -> sin(phi) or cos(phi) will have some random value (between -1;1)* v. Phi value obviously makes no sense anymore. Now I took a model where the coordinate system origin is at the front wheel -> here omega = v / L * sin(delta), so delta -> 90° -> omega = v/L - which is realistic. Then I use coordinate transformation to get back the to rear wheel (as this is my zero point for the agv coordinate system).
@@felix-ke5sr Clearly, you have to use a value of delta that is increasingly close to 90 degrees to evaluate the limit numerically. If you want to do it analytically, you can try to use a power series expansion of cos around 90 degrees.
The kinematic bicycle model is a nonlinear model. Transfer functions only exist for linear models. So there is no transfer function for the kinematic bicycle model. If you are interested in a linear model, with a transfer function, please take a look at the linear bicycle model (in the video on the dynamic bicycle model).
@@professorschildbach First of all, great videos! I seem to have some trouble locating your video with the transfer function, can you link it directly? Thanks in advance.
@@rasm910a Thanks, I'm glad to hear that my work is appreciated! About your question: At the end of the video on the dynamic bicycle model, the transfer function is not stated explicitly, but you can easily derive it from the state space model stated there via the standard formula C(sI-A)^(-1)B. It's a second-order system. We will turn our attention to the stability of that system in due time.
There is a huge number of vehicle models, which are used in all kinds of situations. For example, the dynamic bicycle model (covered later in this course). There are also different kinds of two-track models with increasing degree of complexity.
I have never seen this degree of clarity in my life, not in kinematics at least. I did not want to comment again so as not to annoy, but I had to one more time. Once again Thank you so much.
Thanks a lot ! very clear, it really helped me for a programming project, I'm trying to estimate a car's position in an environnement without GPS.
Thank you for this video. It truly helps me to model the extended Kalman filter.
As Kinematic Bicycle model is applicable at low speed but there is any particularly defined numerical value for speed below which it's applicable.
No. It depends on many factors. It's a model, so it is never 100% accurate. And it tends to be less accurate at higher speeds than at lower speeds.
In 10:22, why not use sin(beta) to calculate R since it is much simpler than using cos(beta)?
There are multiple ways to derive the equations. All of them lead to the same result, of course. It is up to you, which way of calculation you find easier or more intuitive.
Why using the refence point at the middle? Why not only consider the steering angle
Thank you Prof Schildbach for this video. I have a question that we know the kinematic of a bicycle can be written as a nonlinear system like \dot x = f(x,u) and how to check controllability of this system?
what about truck with trailer model?
Thank you Professor, very clear. I have a question with the last step in the derivation. Unfortunately my high school trig is lacking. So to anyone reading this, if you could enlighten me it would be most appreciated (exam's next Tuesday :/).
How do you get from (1/lr+lf)*tan(d)*cos(B) ---to---> B = atan[(lr/lf+lr)*tan(d)] ?
You get there from tan(B)=lr/(lf+lr)*tan(d). Good luck on the exam!
@@professorschildbach Ah ok, so you just used the tan(B) which = lr/R(||) and then substituted in for R(||). The switch from cos to tan threw me for a second. But because the angle B can be expressed in any of the following ways sin(B)=lr/R, or cos(B)= R(||)/R, or tan(B)=lr/R(||) it's all the same. And you needed to find R first that's why used cos(B) instead of going straight to tan.
Thanks for getting back to me
Awesome thank you so much for the time and effort you put into this
Professor, may I ask when to use rear wheel reference and front wheel reference since mostly we use center point reference. Thanks
It is difficult to give a general answer. The the rear axle is typically used as a reference point for low-speed applications (like parking). The CoG is often used as a reference point for high-speed applications. The front wheel is rarely used, in my experience. However, it is used, for example, as the reference point for the Stanley Controller.
wow , very nice explanation, thank you
Percise and clear.It was great.Thanks
Hello, i am wondering does this model has 2 degrees of freedom or 3 ? i cant understand
This model has 3 degrees of freedom: x, y, and psi
It was very usefull to me . Thanks
What if delta gets to 90 degrees?
Kann man die definitionslücke durch wechseln eines Modells umgehen?
First, a normal car will not allow a steering angle of 90 degrees. That being said, the intuitive answer to that question should be clear: The vehicle will rotate around the center point of the rear wheel. Now, what happens mathematically is a little more interesting and requires to calculate the limit of R as delta -> 90 degrees. With the formulas in the video, it should give you l_r, but I have not tried.
@@professorschildbach TY for your response. I am using the model for an agv (one front wheel, steering not compareable to ackermann steering from a car). When I drive the agv with steering 90° it rotates around the rear wheel, just like you said. I am having trouble to model this mathematically.
when delta -> 90° then tan(delta) -> inf => Therefore: omega (angular velocity) -> inf
In my model i integrate omega to get the angle phi. And I use phi to calculate d/dt*x and d/dt*y (both should be zero, I take the rear wheel as origin of the coordinate system) -> as omega is inf -> sin(phi) or cos(phi) will have some random value (between -1;1)* v. Phi value obviously makes no sense anymore.
Now I took a model where the coordinate system origin is at the front wheel -> here omega = v / L * sin(delta), so delta -> 90° -> omega = v/L - which is realistic. Then I use coordinate transformation to get back the to rear wheel (as this is my zero point for the agv coordinate system).
@@felix-ke5sr Clearly, you have to use a value of delta that is increasingly close to 90 degrees to evaluate the limit numerically. If you want to do it analytically, you can try to use a power series expansion of cos around 90 degrees.
Can you make a video with transfer functions for this model?
The kinematic bicycle model is a nonlinear model. Transfer functions only exist for linear models. So there is no transfer function for the kinematic bicycle model. If you are interested in a linear model, with a transfer function, please take a look at the linear bicycle model (in the video on the dynamic bicycle model).
@@professorschildbach First of all, great videos! I seem to have some trouble locating your video with the transfer function, can you link it directly?
Thanks in advance.
@@rasm910a Thanks, I'm glad to hear that my work is appreciated! About your question: At the end of the video on the dynamic bicycle model, the transfer function is not stated explicitly, but you can easily derive it from the state space model stated there via the standard formula C(sI-A)^(-1)B. It's a second-order system. We will turn our attention to the stability of that system in due time.
@@rasm910a Were you able to do it. I understood about A and B in SS format but don't get what is C
@@professorschildbach I get what is A and B but could not understand C, how do we find it sir
Can anyone tell me some models other than bicycle model?
There is a huge number of vehicle models, which are used in all kinds of situations. For example, the dynamic bicycle model (covered later in this course). There are also different kinds of two-track models with increasing degree of complexity.
Thankyou so much for this video