The discussion focuses on calculating the torque produced by brake pads on a hollow-cylinder wheel model. The user seeks to express this torque, initially referring to it as momentum, which leads to clarification that torque is the correct term. Key equations for torque and moment of inertia are shared, emphasizing the relationship between force, distance from the axis, and angular acceleration. The conversation also touches on the effects of static friction and the importance of considering forces acting on the wheel, especially in the context of a vehicle's braking system. The user plans to explore how various factors, including temperature and friction, influence stopping distance.
#181
ROOT0X57B
83
7
jbriggs444 said:
I see you computing the the torque on the tire as:
"couple_freins = -2 * (pression_freins * surface_plaquette) * (rayon_frein / rayon_pneu)"
That appears to be mis-named. It is not a torque. It is a force. Worse, it is not the relevant force.
jack action said:
If you consider the traction limit, the acceleration will always be ##\mu g\frac{M}{M_{eq}}##. Even if you apply a braking torque that doesn't reach that limit, you could simply say you apply, for example, an 80% braking of the maximum value and you would get a deceleration of ##0.8\mu g\frac{M}{M_{eq}}##.
That being said, you have mistakes in your equations.
First - not a mistake per se, but confusing - the equation for couple_freins is not a torque, it's a force. So the unit is N, not N.m.
You both said the same thing, and yes, my bad, that's a misspelled variable
jack action said:
The acceleration in your program is a constant. a = F_eq / M_eq will always be the same, so no need to include it in a loop.
Yes, I know, but because I will add other forces later, I keep it here for consistency
jbriggs444 said:
You multiply the brake pressure (pressure of brake pads on brake disc) by the pad surface area. That part is fine as far as it goes. But it is a calculation of the normal force between pads and disk. You need to multiply this by the coefficient of friction of pad on disc to get a figure for the frictional force between pads and disk. That is an error by a factor of 2.5 to 3.
jack action said:
Second, as I said in an earlier post, the equation for the brake torque is wrong. I still don't know where you got your brake pressure, but I still suspect you are referring to a brake line pressure. This must be multiplied by the total piston area on one side of the brake caliper. For your car, it has a single 48mm diameter piston for the front wheels. For the rear wheels, you have drum brakes with wheel cylinders. The equation I'm presenting to you right now doesn't apply to drum brakes, the actual one is more complicated. Let's assume the rear brakes have comparable disc brakes for now.
That being said, multiplying the brake line pressure by the piston area gives the normal force acting on your brake pads. You still have to multiply it by the friction coefficient of the brake pad (which is usually around 0.38). Then, you multiply by 2 because you have two friction forces acting on your disc: one on each side.
But you also have 4 wheels, so you have to multiply that force by 4 to get the total braking force on your car. (That is really simplified, more on that below.)
I will admit front and rear brakes are the same, I don't want to bother myself with drum brakes
I will multiply the pressure by the total piston area
I should be ##4 * 2 * \pi * 48^2##mm^2 right ? (4 because 4 wheels, 2 because 2 sides)
I forgot the friction coefficient of the brake pad, okay
jack action said:
Weight Transfer
When you are braking, there is a weight transfer from the rear axle to the front axle. This weight transfer depends on the deceleration value. This means that there is a greater normal force on the front wheels compared to the rear wheels. This in turn means that there is a greater friction force available at the front axle than at the rear axle. This means that the rear brakes must provide less braking torque than the front ones. Since the sum of the normal forces acting on each wheel will always be equal to ##Mg##, it is possible to model it as a single big brake that stops the entire car instead (like we did here). But it is worth noting that the friction coefficient of a tire decreases as the normal force increases, so the average friction coefficient of all four tires - with different normal forces - will most likely be smaller than one single tire with the average normal force acting on it.
Hum, I think considering weight transfer effects will complicate my equations so much that it will be very inconvenient...
jbriggs444 said:
What is your estimate for ##A_B## ("surface_plaquette")?
Code:
surface_plaquette = 0.0057390
So that is 0.0058 square meters. 58 square centimeters or about 9 square inches. I have no good way of checking this figure for accuracy. How did you arrive at it?
For this, I took the dimension of the brake pads corresponding to the tire (found https://www.mister-auto.com/plaquette-de-frein/bproauto/pro-0817070/), multiplied length by width, and multiplied by a rough 0.8 because it's smaller than the bounding rectangle
jbriggs444 said:
Note that I disagree with the code in the remainder of the loop. If one is sliding then the moments of inertia of the tires cease to be relevant and the M_eq variable needs to reflect the total mass of the vehicle plus wheels unmodified by their moments of inertia.
Yess, I did not see that coming, I will then change ##M_{eq}## whether the car is sliding or not
According to what I have,
##M## = 830 kg
##M + 4I/{R_\text{ext}}^2 \approx## 830.0.9 kg
Finally, another big error is in your ref_roue_vers_dimensions function. The width of the tire is defined in millimeters, not in centimeters. This gives you an irrealistic value of 0.8 for rayon_frein / rayon_pneu and it should be closer to 0.5.
I think this has more to do with what we said in posts 178 and earlier: I took ##R_B = 0.8R_\text{ext}## but @jbriggs444 told me it should be something like ##0.45R_\text{ext}##
jack action said:
That being said, can you see that the calculations including a single brake pressure, piston area, and disc radius are all rather useless?
Okay, so I will first make all the changes, and then, what can I do with those brakes? It's not like they do not affect the braking distance right?
EDIT: Once I've made all the changing in my code :
[CODE lang="python" title="After modifications" highlight="3, 9,12"]while v > limite_zero:
t = t + dt
force_freins = -2 * 4 * mu_plaquettes * (pression_freins * surface_piston) * (rayon_frein / rayon_pneu)
...[/CODE]
With the same technique as before (changing the pressure and running the script again), I find a limit of 5 bar before sliding
I think I still have to modify something
Hum, I think considering weight transfer effects will complicate my equations so much that it will be very inconvenient...
All I want you to see is that the calculations about the torque coming out of the brake system are irrelevant to your problem and other stuff might have more impact.
ROOT0X57B said:
I think this has more to do with what we said in posts 178 and earlier: I took ##R_B = 0.8R_\text{ext}## but @jbriggs444 told me it should be something like ##0.45R_\text{ext}##
No, you have an error in that function that affects greatly your radiuses:
Python:
def ref_roue_vers_dimensions(roue):
"""
Convertit une référence de pneu en un rayon et une largeur
Les références sont du type A-BRC avec
A : Largeur de la roue en cm
B : Ratio hauteur de la bande/largeur
C : Diamètre de la jante en pouces
"""
largeur_cm, infos = roue.split('-')
ratio, diametre_jante = infos.split('R')
largeur = float(largeur_cm) / 100
diametre_jante = float(diametre_jante) / 0.0254
epaisseur = largeur * float(ratio)/100
rayon = diametre_jante / 2 + epaisseur
rayon_interieur = diametre_jante / 2
It should be:
Python:
def ref_roue_vers_dimensions(roue):
"""
Convertit une référence de pneu en un rayon et une largeur
Les références sont du type A-BRC avec
A : Largeur de la roue en cm
B : Ratio hauteur de la bande/largeur
C : Diamètre de la jante en pouces
"""
largeur_mm, infos = roue.split('-')
ratio, diametre_jante = infos.split('R')
largeur = float(largeur_mm) / 1000
diametre_jante = float(diametre_jante) * 0.0254
epaisseur = largeur * float(ratio)/100
rayon = diametre_jante / 2 + epaisseur
rayon_interieur = diametre_jante / 2
Your tire is not 155 cm wide, it is 155 mm wide. With your code, you have a tire width of 1.55 m, a wheel diameter of 551.18 m (I missed that mistake), a sidewall height of 1.0075 m, and a tire radius of 276.6 m. This gives you rayon_frein / rayon_pneu = (0.8 * 275.59) / 276.6 = 0.797.
The correct numbers would be a tire width of 0.155 m, a wheel diameter of 0.3556 m, a sidewall height of 0.10075 m, and a tire radius of 0.27855 m. Resulting in a rayon_frein / rayon_pneu = (0.8 * 0.1778) / 0.27855 = 0.51.
And you are also using this ridiculous tire radius to evaluate the equivalent mass.
ROOT0X57B said:
Okay, so I will first make all the changes, and then, what can I do with those brakes? It's not like they do not affect the braking distance right?
They do not affect the braking distance. Let me repeat that such that it sinks in:
I see that we are now talking about pressure in the brake lines and surface area of the pistons.
I do not know how reliable the 14 mm figure is for piston radius. It seems larger (28 mm diameter) than I would have expected for a small car brake cylinder. So am not confident in the correctness of the "surface_piston" variable. In the case of four wheel disc brakes, I would expect larger diameter cylinders in front and smaller in back.
It would be good if, rather than presenting a program and reading the numerical result from a set of output graphs one used algebra and produced a formula and a calculation using that formula.
If we set force_freins to limit_glissement, that gives us an equation that we can solve for pression_freins.
Let us change up the variables a bit for the shift from Python to LateX.
force_freins = ##F_\text{tire}## The braking force deliverable by the tire to the road in the absence of slip.
limite_glissement = ##F_\text{road}## The limit imposed by the friction of tire on road.
mu_plaquettes = ##\mu_\text{pad}## The coefficient of friction of pad on disc
pression_freins = ##P_\text{line}## The pressure of the brake fluid in the brake lines.
surface_piston = ##A_p## The area of one brake piston.
rayon_frein = ##R_B## The radius to the [middle of the] swept area of brake pad on disc
rayon_pneu = ##R_\text{ext}## The radius of the tire.
mu_sol = ##mu_\text{sol}## The coefficient of friction of tire on road
masse = ##M## The total vehicle mass
g = ##g## The standard acceleration of gravity, 9.8 m/s^2
$$ 8 \mu_B P_\text{line} A_p \frac{R_B}{R_\text{ext}} = F_\text{tire} = F_\text{road} = \mu_\text{sol} M g$$We are solving for ##P_\text{line}## so we divide by everything else on the left hand side. That yields:
$$P_\text{line} = \frac {R_\text{ext}} {R_\text{B}} \frac{\mu_\text{sol} M g} {8 \mu_B A_p}$$That result looks right. It is directly proportional to the things it should be and inversely proportional to the things it should be.
You want to know why that result is equal to 5 bars. If it is equal to 5 bars, it is so because of the inputs that went into the calculation.
Where does that value come from? I got 48 mm diameter from the source in the link (Diamètre du piston d'étrier de frein: Caliper piston diameter):
The 14 mm value appears to be my typo for the correct value of 24 mm since @ROOT0X57B is using the figure in a formula for area: ##a = \pi r^2##.
So that could be a factor of 4 error in the result -- if we were shown a calculation for the piston surface area.
[VENT]Would it be so very hard to actually get all of the inputs and calculations together in one post without burying them hip deep in some Python code[/VENT]