Best way to slow down.

I’ll try to bring some light into this discussion (and confusion) by applying the appropriate physics.

The theory, picture and formula about braking and brake balance is from the book “Rennwagentechnik” (race car technology) by Michael Trzesniowski. Sorry, it looks like it is available only in German and that there is no English version. So I will try to translate the most important things. The book can found here Rennwagentechnik: Grundlagen, Konstruktion, Komponenten, Systeme (ATZ/MTZ-Fachbuch) (German Edition): Michael Trzesniowski: 9783834817792: Amazon.com: Books
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First let’s have a look at the physics of a car when it is braking (see pic below).

L(f) = Distance from center of gravity of the car to the front axle
L(r) = Distance from center of gravity of the car to the rear axle
h(v) = Height of the center of gravity above ground (road)

The braking force applies at the wheels were they make contact with the road. See F(W, X, B, f) for front and F(W, X, B, r) for rear.
The load of the axels is F(V, Z, f) for the front and F(V,Z, r) for the rear axle.
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Now let’s have a look at the formula. It looks more complicated than it is.

The formula shows the axle load distribution between the front and the rear axle.
a(x) = Braking force that “attacks” the center of gravity of the car
g = gravity (9.81 m/s)
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To make things easier to understand, let’s assume we don’t brake and thus have no brake force (a(x) = 0). This results in the formula below which says the axle load distribution between the front and the rear axle is according to where the center of gravity is between the front and rear axle. If it is in the middle (L(f) = L(r)) then both axles are equally loaded. If the center of gravity is closer to the front axle, the front axle is more loaded than the rear axle according to the proportion of L(f) and L(r).
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Now let’s see what happens when we are braking (a(x) > 0). Now the second terms in the formula below are different from zero (h(v) * a(x) / g).
So the formula says that in addition to the load of the static car there is load generated by the braking force. This adds load to the front axle and “removes” load from the rear axle. The formula says, the harder you break (a(x) >> 0) and the higher the center of gravity (h(v)) the more load on the front axle and the less load on the rear axle.

 

You can also see that there is a point when there is all load on the front axle and no load on the rear axle. This is when the denominator is zero:

l(f) = h(v) * a(x) / g

This means that a short car (l(f)) with a higher center of gravity (h(v)) loses grip on the rear axle earlier than a long car with a lower center of gravity. Just imagine a 100m high car which is only 3m long (well, it’s more a tower than a car then). It doesn’t need much breaking power to flip this car.

 

Now this is where the example with the bike comes into play. 4th_gear mentioned that you can flip a bike when applying the front brakes hardly, but that you cannot flip it by the rear brakes.

The point where you flip the bike by the front brake is when l(f) = h(v) * a(x) / g. So you need a high center of gravity h(v) on a short bike and strong enough front brakes to easily flip a bike.

BTW you could flip a car too with strong brakes. The only difference of a car over a bike is that the center of gravity is much lower compared to the length of the car/bike. This is why you can’t flip a car just by the front brakes. But think about my high and short car again (the “tower”). It would be very easy to flip such a “car”.

 

Why can’t you flip a bike (or car) with the rear brakes?

Actually, you could flip it too….. but there is one “problem”. As we have seen, the weight gets transferred to the front axle when braking (no matter if you brake with the front or rear brakes). So let’s brake hard with the rear brakes so that we will lose grip at the rear axle (since all weight is now on the front axle). What happens? The rear axle loses grip and thus is not able to transfer any braking force (plus the car would spin…). So you never can apply as much braking force by the rear axle because the more you brake the “lighter” the rear axle gets thus not being able to transfer the braking power (grip).

As soon as the car would start to flip, you will lose braking power on the rear axle, thus preventing the car from flipping.

 

Even if the rear wheels had infinite grip you couldn't flip the car because the rear axle is acting like the pivot.

With virtual infinite grip, during hard braking, the cog of the car would just tend to get on an horizontal line in front of the braking axle. So it would flip only if the front axle is braking too and the cog is higher than the front axle. If the cog is lower than the axles, the car would pitch upwards instead of dive downwards

I wish I had studied physics. I love this stuff

 

According to the book, the braking power of the front brakes is about 65-75% whereas the rear brakes only have (the remaining) 25-35%.

This is why the front brakes are stressed much more than the rear brakes and why the front brakes are much more important for braking that the rear brakes.

Plus there is another very important thing: What happens if your front tires get blocked because you don’t have ABS and press the brake pedal very hard? You cannot steer the car anymore (until you release the brakes as far as the tires don’t block). Besides that nothing happens to the car (well there is some smoke coming from the tires some when…).

What happens if the rear tires block due to too much braking? The car spins because it has lost its stability which comes from the rear axle.

So if you use your engine to “improve” braking power on the rear axle you are playing a very dangerous game. I guess most of you (including me) had lost your car when being hard on the brakes and downshifting at the same time without proper revs (heel and toe).

 

Agreed.

 

Gentlemen, I appreciate your taking the time to share your thoughts and explain your views on the matter. We all know the front brakes are much more powerful than the rear brakes and I appreciate Elsi providing the comparative amounts of braking normally delivered by the front and rear brakes. They are what I expected.

I am having another busier day and have a guest coming over today as well. Let me first reply to the points raised about braking in a (race) car as they address the OP's question. I'll explore what actually happens in engine braking as well as rear braking on a bike separately.

SLOWING vs. STOPPING
With regard to a (race) car braking, I think we need to determine if we are simply slowing the car or doing threshold braking. If we are just slowing the car, as in a fast corner, we don't need or want maximum braking whereas if we are braking to stop the car or navigate a very tight hairpin, we do want maximum braking.

So when we are just slowing to take a corner at the highest speed, we want sufficient braking, superior control and the best setup for corner exit speed. The issue of balance is critical in achieving the best control and setup for the fastest corner exit. The front brakes may have the best grip but if we have too much front brake bias we end up with understeer.

I have indicated this since my first post on this thread... it's desirable to modulate brake bias with engine-braking to achieve the best balance. While front brakes do provide most of the stopping, it is not always the most important or the only goal under all racing conditions. Please keep in mind that in racing or just going fast, you don't actually want to stop. You just want to quickly get in the right gear and engine speed for a given maneuver. IMO engine braking modulates the effects of front brake bias results in a better weight distribution on the chassis when cornering.

If you go back and listen to the XX drivers go around the track at Ferrari sporting venues, they downshift continuously as they go around long fast corners. They do not just brake and downshift once.

 

No need to hurry, 4th_gear.

I just wanted to compile the physics and post them here. I was curious by myself how this really works, so I studied the book I mentioned and compiled the results of my studies.

Enjoy driving your car and be careful on the brakes and with the motor

Markus

 

Thanks Markus. Just like you and others here, I often find it fascinating how physics explains the handling behaviour of our babies.

 

You're referring to a post I made 2 nights ago. I was being honest and trying to see if you would prove me wrong.

You posted a short parting message, said I was confusing differential loading with engine braking, and spent no effort to explain that assertion. That was impolite. Then you also threw in very badly mis-worded parting remark about the 599's cornering. It looked to me you just wanted to back out and throw a monkey wrench into the discussion on your way out. If you really had something to say about engine braking you could have said it in the 2 days that elapsed since I wrote my reply.

Now 2 days later, you haul out the big names "Skip,Barber or Bertil Roos" (again). BTW, it's "Skip Barber", just one person, not 2 persons; not "Skip, Barber". Yeah, so you've been following discussions that I participated in elsewhere. What other out-of-context material do would you like to add to this thread?

BTW, I believe weight does not transfer to the front (axle) when you brake - IT SIMPLY TRANSFERS TO (PIVOTS OVER) WHERE THE BRAKING IS BEING APPLIED . So if you have 100% rear brake bias, weight transfers to the rear axle... but because the car is a rigid structure and more of the car is in front of the rear axle, a larger portion of the car's overall weight also "transfers" against the front axle.

 

Weight, generally speaking, is aligned to the gravitational vector (vertically). There is no weight transfer due to braking, i.e. mass is not redistributing. The increased loading that occurs to the rear axle in your example above is the balancing force for the inertia. Since the car can not rotate about this pivot pt (rear wheel contact with the ground) there is required a balancing couple between front and rear wheels. This balancing couple results in a reaction force oriented upwards on the front and downwards on the rear. Therefore the resultant vertical load component (weight + inertia reaction force) on the rear wheels decreases while it increases on the front. The figure shown in post #52 explains it. The same holds true whether you are using the brakes or engine to slow down, i.e. the horizontal line of action of using the brakes (front or rear) is same as for engine braking.

 

I'm here."heel&toe"

 

Weight aligned with the "gravitational vector" WTF? It's not rocket science, and we're not landing on Mars. Brakes are for slowing down. Gears are for keeping the engine within the power band. Period.

 

Thanks for everyone's input. I'll try to understand your points and hopefully comment on them before too long. I was kept quite busy at home over the last 2 days.

I'm should also catch up on sleep but will post a few drawings to let you think over them. I modified my original drawings regarding my bike braking" example to account for some newer insights on the complexity of what happens if you suddenly stop the bike by arresting the motion of the rear wheel on the bike.

I started by locating a drawing of a cyclist on a bike with COG kindly labeled by the unknown Internet author. I converted this to a stick-diagram.

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I then tried to identify the "forces" acting on the bike

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I then show the bike as a single rigid object with one COG vs. the bike as 2 separate objects with separate COGs

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I then analyzed and show the bike as 2 objects hypothetically linearly stopped by a "giant hand" gripping the rear wheel where:

A = pivot position relative to COGs at initial braking
B = imaginary "mid-air" visualization of objects acting on separate pivots as cyclist pivots at the handlebar
C = simulation of objects responding on a road with angular momentum acting along separate radii to separate pivoting points

Important equation for cyclist is SPEED = ANGULAR VELOCITY x RADIUS

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This drawing needs more work to flesh out than the others. It shows the bike as a single rigid object hypothetically linearly stopped by a "giant hand" gripping the rear wheel where:

A = pivot position relative to the COG at initial braking
B = imaginary "mid-air" visualization of the object spinning as linear speed is converted to angular velocity over the radius from the pivot
C = simulation of the object responding on a road with angular momentum acting at 2 points of contact on the road surface. The original momentum of the object is probably expended as heat at the braking "hand" and 2 points of contact on the road.

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These drawings are just my semi-educated guesswork. I'll try to refine/correct the last 2 drawings when I have more time. I would welcome constructive comments.

ENGINE-BRAKING
Finally, as for how engine-braking might be analyzed in a similar fashion, I'm somewhat challenged when considering the actual workings of engine-braking as it's created by the resistance from uncombusted air intake in the combustion chambers, retarding the spinning of the crankshaft, at the engine. So if you equate the compressed air resistance in the combustion chamber to the "imaginary hand" I used in the bike example, can you say engine braking occurs at the rear axle at all?

Good night.

 

The giant hand does not grip the rear wheel at the location shown. Nor is there a downward acting force at that location. See the figure in Post #52. Forget about any internal forces in the system, e.g. engine , occupants. They do not have any effect on the external force balance. You are just confusing the picture and yourself.

 

I was going to say:
Correct the FBD. The forces can only act through the tire contact patches. Remove the Hand Of God and replace it with force vectors acting against the direction of travel pushing through the contact patch. Label them FBF and FBR for Force Braking Front & Rear. Now add the weight on each tire. Label them FNF and FNR for Force Normal Front & Rear.
Get rid of the stick guy. While he's amusing, he's distracting. Just give our vehicle a center of gravity.
Now calculate the moment about the CG from a braking force acting entirely at the front wheel, and one acting entirely at the rear wheel.

P.S.- the answer to where to consider the engine braking acting is at the beginning of this message.

 

Sorry, I had wrongly assumed all readers would have read and kept in mind ALL the responses from people posting on this thread. That was silly. My mistake.

The "giant hand" (of God ) was a response and my interpretation of a couple of posts Igor Ound's post about the bike analogy. He wrote "Imagining as an extreme a virtual instant braking force like a rope attached respectively only on the front or the rear axle...".

He also wrote an earlier post that I said I would reply to where he said "If you attach a rope from a wall to the rear wheel and ride fast until it tenses, you will definitely go over the bars. Or if the rear wheel gets trapped in a manhole".

I preferred and used the "giant hand" because I had already mentioned the giant hand analogy in an even earlier post, where I wrote "Imagine the whole car dangling in the air being supported only by a big hand with thumb and fingers holding the drive axle.". You'll find it in the 3rd last paragraph in this post.​

I apologize for the oversight. I felt Igor Ound's reference to my bike analogy was the most interesting suggestion that would benefit from a hypothetical analysis to flush out the forces that act on the bike and cyclist.

Many real world problems can be better understood by using unreal, extreme examples that amplify the action of factors involved in real world situations. Scientists do that to reveal their presence and this allows better understanding of the whole system of factors that make up the final picture in the real world.

Please keep in mind, it's easy to criticize in a discussion but what makes the difference is if you can offer a credible alternative theory and explain it in adequate detail.

 

jcurry, I appreciated, largely understood and agreed with your earlier comments in Post #66. I still have to sort through the complete picture but your comments and Elsi's helped a lot.

However, as for the effects of the engine from engine braking, I am still not clear on this. When the engine is coupled to the contact patches, both ends of the interaction must simultaneously be affected when the engine is quickly braked. My assertion is that the engine is quite a distance away, with its own rotating masses (it appears to me the engine is not rigid with the rest of the car) so we should account for what's happening at the engine's location.

 

Yes, aside from wind resistance, the braking forces have to act on the contact patches. However, it gets more complicated when you are no longer braking to stop in a straight line. When cornering there are different and desirable ways of acting on the contact patches, here are some that quickly come to mind:

- changing the weight balance on the patches, especially front vs rear for better rear traction, especially for corner exit and overall control

- balancing the differential rates of retardation of the front tires vs rear tires against the contact patches (e.g. if the front tires brake harder than the rear) this affects the braking efficiency of the front vs rear contact patches

- during cornering where there are angular forces on the patches, increasing slip angles of the rear tires to reduce understeer caused by front brake bias

- using throttle in a lower gear to manipulate the slip angles of the rear tires to control the balance of understeer/oversteer (you can't use throttle if you are applying the brakes) but you can downshift and get back on throttle and manipulate throttle to brake the engine as well as change the slip angles

- in a long race, you may also want to conserve your brakes by spreading the work​

 

Dude,
seriously. We were almost making progress analyzing the bicycle in a straight line.
Let's keep the analysis simple and eventually we may get everyone on the same page here rather than throwing a lot of red herrings back into it.

Analyze the bicycle in a straight line with the free body diagram corrected. Calculate the moment about the CG from front-only and rear-only and feel free to do a few mixtures of the two.