3 жыл бұрын
3 жыл бұрын
3 жыл бұрын
3 жыл бұрын
3 жыл бұрын
3 жыл бұрын
Пікірлер
@anonymousKern 14 күн бұрын
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 14 күн бұрын
You get there from tan(B)=lr/(lf+lr)*tan(d). Good luck on the exam!
@anonymousKern 14 күн бұрын
@@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
@Alithenius Ай бұрын
I'm a math student and somewhat new to this engineering content. At around 6:40, would it be correct to say that the path frame will be the usual tangent, normal, binormal vector frame one would get from vector calculus? The d-axis being the tangent vector, e-axis being the normal vector, and n being the binormal vector?
@professorschildbach Ай бұрын
Yes, that would be correct.
@slyguy8808 2 ай бұрын
I am actually studying in another city, and we do not have this kind of high quality course on Vehicle Dynamics, but I need some knowledge about it for a big project. Therefore, I am really glad that you are sharing this course.
@user-po1zk3yp4c 2 ай бұрын
clear and easy Thank u
@rabbigolam4789 3 ай бұрын
Prof. can I please have the notes? They are amazing.
@bjj-vl2ut 3 ай бұрын
FWD cars understeer because they have more weight over the front, but lift-off oversteer occurs because the car's weight transfers forward.
@Waka-bm4ce 3 ай бұрын
Thank you for these courses, really well explained
@user-fo8fn6pp7w 4 ай бұрын
I think if the gear ratio greater than 1, means the speed of wheels slower than the shaft; if less than 1, means the speed of wheels faster than the shaft. However, that is different what professor said ih the lecture
@isleofauto 4 ай бұрын
Hi Prof, excellent explanation! Thank you so much! However, why the anti dive is only analyzed on the front wheel not the rear? What about using the rear-axle only hand brake to brake the car? I did a similar calculation using your anti squat example with rear wheel drive model and got the rear axle weight loss. Is that always the case? A driving instructor once told me using rear brake will "tuck" the car in, I guess he meant squat. But that will by no means happen, right?
@user-kp4iy6sr5n 4 ай бұрын
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?
@Channel_1728 5 ай бұрын
That was great and I agree with the previous comment. And I like the notion of adjusting the pivot points for a more enjoyable driving experience for a standard vehicle. So, if I am a little below and above the pivot points respectively, am still OK unless I am doing a dragster. And that would be its own driving experience LOL.
@jingxuanliu3758 5 ай бұрын
Professor, may I ask when to use rear wheel reference and front wheel reference since mostly we use center point reference. Thanks
@professorschildbach 5 ай бұрын
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.
@back2back135 6 ай бұрын
Sehr gutes Video, dankeschön.
@back2back135 6 ай бұрын
Hallo Herr Schildbach, vielen Dank fürs Hochladen der tollen Videos. Gibt es für die anderen Aufgaben vom Aufgabenblatt 01 auch Musterlösungen? Falls nicht, könnten Sie die bitte auch Hochladen? Auch wenn es nur die Lösungen ohne Rechenweg gibt, wäre ich froh diese zu erhalten.
@back2back135 6 ай бұрын
Danke fürs Hochladen.
@back2back135 6 ай бұрын
Eine Frage: Im Buch "Technische Mechanik 1" von Gross steht "Die Richtung der Kraft können wir durch ihre Wirkungslinie und den Richtungssinn auf ihr beschreiben." Das heißt, Richtung = Wirkungslinie + Richtungssinn . Die Vektoren bei 4:24 haben dieselbe Wirkungslinie aber einen verschiedenen Richtungssinn deshalb ist die Richtung der beiden Vektoren verschieden. Oder sehe ich da was falsch?🤔🤔 Es scheint, dass bei Ihnen mit "Richtung" das gemeint ist was in Gross mit "Wirkungslinie" bezeichnet wird.
@professorschildbach 6 ай бұрын
Danke für den Hinweis. Wir verwenden hier die Terminologie aus dem Buch "Technische Mechanik 1" von Hagedorn und Wallaschek. In dem von Ihnen genannten Buch wird eine leicht andere Terminologie verwendet. Wichtig ist nur, dass die Terminologie konsistent ist, damit die Bedeutung immer klar und eindeutig ist. Und Sie sollten natürlich die Terminologie aus Ihrem Kurs verwenden.
@back2back135 6 ай бұрын
@@professorschildbach Danke für die Antwort und für das Hochladen der Videos.
@back2back135 6 ай бұрын
Danke fürs Hochladen!!!
@nehadave3877 6 ай бұрын
What is my cars front n rear trackwidth is different?..how will I calculate the roll angle..and what will be the value of "b"?
@a.n.7761 7 ай бұрын
Wo bleibt Tm3 😊?
@SendItStephen 7 ай бұрын
How would you work out a pivot point for a trailing arm suspension?
@boi829 8 ай бұрын
this is the best explanation of anti-dive and anti-squat i've seen, just because it's in terms of moments and not just geometry (and trust).
@raafeh9601 8 ай бұрын
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.
@raafeh9601 8 ай бұрын
Thank you so much for such a clear presentation of fundamentals. These lectures are a gem.
@user-lw4fo9nt3j 8 ай бұрын
could you recommend some relative textbooks, which contain more details like explaining the induction of the equations in your video? thanks a lot.
@richardchaney7295 8 ай бұрын
It appears to me that, for a 3-dimensional rigid body, the instantaneous centre of rotation is not a single unique point, but rather all points along some line.
@professorschildbach 8 ай бұрын
Yes, the instantaneous center of rotation only exists for planar (or quasi-planar) motion.
@seongyunkim9487 9 ай бұрын
I have a question regarding a research paper I found dealing with vehicle dynamics in oversteer. It deals with elliptical tire models and load transfer, and I can’t understand some of the substitutions the writer makes into the equations of motion. I already sent you an email named “Vehicle dynamics question” through your university email address. It would be very helpful if you can reply to it. I really enjoy your videos
@professorschildbach 9 ай бұрын
I am glad you like the subject and you enjoy my videos. Please understand that I lack the time to support individual projects or explain details in papers which are not covered in my lectures. Best of luck with your project!
@user-jt2zr8wy5t 10 ай бұрын
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 10 ай бұрын
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.
@KenW0728 11 ай бұрын
Really good lecture. Can I translate it into Chinese and repost it on bilibili? I'll mark your name and this youtube address there
@professorschildbach 11 ай бұрын
Yes, you can. Thanks for asking.
@KenW0728 11 ай бұрын
@@professorschildbach Thank you sir.
@tobiasfischer6755 11 ай бұрын
7:45 should be 1 rad/s = 30/pi rpm
@professorschildbach 11 ай бұрын
Right. Thank you for your this correction.
@catalinagonzalez5498 Жыл бұрын
genio
@badrboutara574 Жыл бұрын
Bitte warum gibt es ein Minus beim omegazwei ( W2)
@professorschildbach Жыл бұрын
Über welche Stelle im Video reden wir hier?
@evanparshall1323 Жыл бұрын
Wow this is a great explanation!
@k4piii Жыл бұрын
what a master class! I can not believe this is free! but x1.75 is the way to go xD
@berndmayer3984 Жыл бұрын
der Ort von B(t) lässt sich geometrisch leicht als Summe zweier cosinusse darstellen. Und dann 2 mal nach t differenzieren. Jedenfalls meine "Physikerlösung" (ohne Vektoren und Kreuzprodukten) falls sonst nichts weiteres gesucht wird. Das video ist sehr allgemein aufgebaut und zudem speziell technische Mechanik.
@professorschildbach Жыл бұрын
Ihr Ansatz ist selbstverständlich auch korrekt und führt auch zum Ergebnis. In diesem Beispiel geht es aber nicht nur darum, das spezifische Problem zu lösen. Es geht um die Lehre einer allgemeinen Methodik, in diesem Fall die der graphischen Lösung. Graphische Lösungen haben gegenüber einer Rechnung Vor- und Nachteile. Der wichtigste Vorteil ist die Veranschaulichung, in diesem Fall der einzelnen Geschwindigkeits- bzw. Beschleunigungskomponenten. Ich lege bei meiner Vorlesung immer Wert auf Rechnung UND geometrische Anschauung, und insbesondere die Fähigkeit, beides zu verbinden. Einen Unterschied zwischen "Physikerlösung" und "Ingenieurslösung" kann ich grundsätzlich nirgends erkennen. Alle korrekten Lösungen sind valide. Hier geht es eher um die Unterscheidung zwischen einer "problem-spezifischen Lösung" und einer "allgemeinen Lösung". Mit Vektoren und Kreuzprodukten haben Sie auf jeden Fall ein sehr allgemeines Werkzeug an der Hand. Aus meiner Sicht lohnt es sich, dieses Werkzeug zu beherrschen. Es gibt definitiv Problemstellungen, in denen Sie mit einem reinen trigonometrischen Ansatz in grosse Schwierigkeiten kommen.
@emiliolozanolozano7186 Жыл бұрын
Thank you Georg. I can see in the video that, when explaining the antisquat on acceleration, you consider the force reaction at the wheel contact point with the ground. Why do you think Miliken or Dixon consider this force applied to the wheel center? If you could recomend a more detailed explanation with the analysis of the forces on teh suspension elements would be very much apreciated.
@thinkstorm Жыл бұрын
I was wondering the same thing! The load transfer happens at the contact patch, but the torque from the engine->transmission->diff->IRS axles would happen at the wheel bearing... I'm confused.
@dielaughing73 6 ай бұрын
@@thinkstormtorque isn't really localised like that. Any moment acting at the wheel bearing would be in the form of a frictional loss.. the torque is applied to the axle, but transmitted at the road
@setsotothebenyane8079 Жыл бұрын
Great lecture series, absolutely beautiful but is this the last video?, it doesn't feel like the video
@setsotothebenyane8079 Жыл бұрын
Great lecture series, absolutely beautiful but is this the last video? It didn't feel like the last class
@larsprokopetz4476 Жыл бұрын
Sehr gutes Video, sollte mehr als genügend aus reichen, für mein Kurz Test morgen
@SoumilSahu Жыл бұрын
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 Жыл бұрын
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.
@adamelwood7768 Жыл бұрын
Could the "VPP" be considered the vehicles center of gravity?
@tesie8849 Жыл бұрын
Unglaublich hilfreich gewesen, vielen Dank 😊
@zaidmohammadal_rababah6439 Жыл бұрын
Prof, may i ask you which books are recommended for vehicle dynamics because i am looking for a deep knowledge in the field to make a project of it. your help is appreciated
@Gnarkson Жыл бұрын
Persönliches Inhaltsverzeichnis: 07:16 - Lager / Auflager
@kamil1088 Жыл бұрын
Guten Tag Hr. Prof.G. Schildbach , Sie geben an , dass sich die Waage im statischen Gleichgewicht befindet, wie kann das aber sein wenn in 4:53 Min. gezeigt Fi,x=0 ABER Fi,y=2mg ist. Wenn m Nicht = 0 ist , ist Fi,y Nicht 0 und damit wäre die Wage nicht im Gleichgewicht. Ich bitte freundlich um Feedback. Vielen Dank für die Bereitstellung der Videos zu TM . Besten Gruß Kamil B.
@professorschildbach Жыл бұрын
Die Kräftegleichung wird so aufgestellt, dass Summe F_{i,y}=0 sein muss. Das wird auch erfüllt, und zwar genau dann wenn gilt: L_y=2mg .
@kamil1088 Жыл бұрын
@@professorschildbach Guten Tag Hr. Prof. Schildbach , vielen Dank für die rasche Antwort. Ich werde mir das Video noch einmal anschauen und jenen Abschnitt genauer reflektieren . Besten Gruß Kamil B.
@kamil1088 Жыл бұрын
@@professorschildbach Vielen Dank. Es handelte sich bei mir um einen Gedankenfehler. Vielen Dank und besten Gruß
@cmcpros7403 Жыл бұрын
Delta FC R? Longitudinal tire force?...plain English plz
@user-xn5uu2pc1x Жыл бұрын
How good and useful lecture it is!
@danielbadel1226 Жыл бұрын
Another question if i may ask for the suspension kinematics. For double wishbone suspension on the front and rear, Does the behavior should be similar? I mean the camber gain rate and roll center dynamic position for suspension deflexion and roll angle variation have to be similar on the rear and front axle or should be different?
@danielbadel1226 Жыл бұрын
Hi! Excellent video! I have one question related to the roll axis slope. According to these equations the front and rear roll center hight affect roll moment and thus the roll angle. I may ask if the slope should be inclined upward to the front for RWD and to the rear for FWD? Trying equal contributions lf*hr and lr*hf to the moment assuming in RWD lr is shorter (Heavier on the rear) and FWD lf is shorter. Or which criteria can we use to establish the roll center slope? Thank you for your video and i hope the question is clear
@prabanjan8140 Жыл бұрын
THANKS