Brake System Design and Optimization: Calculating Braking Capacity
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Calculating Braking Capacity
12.58
| 00:00 | - In this section of the course, we're going to be looking at designing and speccing a full brake system from scratch. |
| 00:07 | This process does involve a lot of maths but we're going to be working through it slowly in the following modules to ensure you have a full understanding of the calculations, how they work and why we're doing them. |
| 00:18 | With that said, we've also developed a full brake system calculator which will do all the heavy lifting for you so don't get scared off. |
| 00:27 | Even when you're using the calculator, we believe it's still important to know what's going on in the background which is what you're about to learn. |
| 00:34 | Feel free to jump ahead and have a play around with the calculator at any stage but if you want to make the most out of it and ensure you're getting reliable results, you'll need to pick up the skills we'll be covering over the next few modules. |
| 00:48 | So with all that in mind, let's start learning about how to calculate our brake capacity. |
| 00:53 | To be clear, there's nothing necessarily wrong with upgrading brake parts as you go but if you have the option and the funds allow, I guarantee you'll be better off and thank yourself later if you start from scratch. |
| 01:05 | Figure out exactly what your requirements are, and consider all your parts together as a package to meet those requirements. |
| 01:13 | With each factory component presenting its own set of issues, it's easy to see how it can be worth starting from scratch with a clean slate, rather than working around the constraints of the original system or upgrading components bit by bit when you're probably just going to change it all anyway. |
| 01:30 | Or if we are retaining factory components, it's good to understand what this means for the rest of the system. |
| 01:37 | Before we start, you should know that the following calculations are applicable with or without ABS. |
| 01:42 | Even with ABS, we still want to design our system with the same ideas in mind as the better our system works alone, the less the ABS will need to work and the better the braking performance will be. |
| 01:54 | Throughout these calculations we'll be working mostly with metric units, as long as you understand the underlying ideas, you'll be able to complete the process in either units though. |
| 02:05 | So the first thing we need to consider is the grip available at each axle. |
| 02:09 | As we've discussed in the earlier brake bias and longitudinal tyre force module, we're trying to match the brake torque supplied by the system to the braking capacity of each axle which is a function of the grip available at that axle or rather the longitudinal tyre forces that can be generated. |
| 02:27 | We want to avoid over braking the vehicle where we have lots of braking power but not enough grip as we'll just end up with brakes that are difficult to modulate and lock too easily. |
| 02:38 | If we roughly match the brake torque to the grip available, we can achieve good modulation with predictable engagement to allow for maximum braking without sudden lock ups. |
| 02:48 | This doesn't mean that our brake torque will be limited to our braking capacity. |
| 02:53 | For the coming calculations, we'll set a target brake pedal effort to achieve the maximum braking deceleration. |
| 03:00 | But we'll still be able to apply the pedal force over and above this target which is going to increase the braking torque above the grip available and lock the wheels. |
| 03:10 | Again we just don't want to make this too easy. |
| 03:13 | Calculating the longitudinal tyre force is the tricky bit since it involves a lot of variables, thanks to changing conditions in different tyres. |
| 03:21 | Again as we talked about earlier, the reality is that tyre forces are extremely complex and would require their own course to cover so moving forward in this module, we're going to use a very simplified view of longitudinal tyre forces generated in a braking event. |
| 03:38 | In the brake bias module, of the fundamentals section of this course, we also covered how the vertical load through the tyre, AKA the normal force between the tyre and the road, is a function of the static weight distribution and longitudinal load transfer. |
| 03:54 | So we'll need to know these in order to work towards understanding our brake bias requirements. |
| 04:00 | The best way to determine the static weight distribution is of course weighing the car on corner weight scales. |
| 04:07 | And let's say for example that our car weighs 1200kg and has a 53% front weight bias. |
| 04:14 | So 636kg on the front axle and 564 on the rear axle. |
| 04:21 | The longitudinal load transfer can be calculated with the load transferred off the rear axle and onto the front axle as equal to the longitudinal acceleration or deceleration from braking, multiplied by the total weight of the car and the height of the centre of gravity, all divided by the wheel base. |
| 04:40 | The slightly vague but is the longitudinal acceleration due to braking. |
| 04:44 | The maximum deceleration which is measured in G force will determine how much load is transferred from the rear to the front axle and this will determine our target brake bias. |
| 04:57 | On top of this, when the brakes are applied, how quickly this G force ramps up and also tapers off, depends on the vehicle's speed as well as the suspension setup, tyres and aero. |
| 05:08 | This added complexity is again outside the scope of this course and the only real method of dealing with how the braking capacity changes with these factors is through the driver inputs. |
| 05:21 | So it's reasonable to assume a constant rate of deceleration once the brakes have been applied for a full power stop. |
| 05:28 | We want this to be representative of a maximum braking power stop in ideal conditions and we'll use the bias bar adjustments and a proportioning valve to account for anything less. |
| 05:39 | The key question is, if we're designing a system from scratch, how do we know what it'll be capable of. |
| 05:46 | We clearly need to make some educated assumptions here based on what's realistic to expect given the vehicle's setup and application but most importantly, what tyres it's running. |
| 05:59 | A modern production car on decent tyres might be in the vicinity of 0.8G whereas a performance focused car on suitable tyres could see up to 1G. |
| 06:10 | Well sorted race cars with minimal downforce are able to achieve between 1 and 2G and adding more downforce into the mix could bring this up to 3G and just for interest's sake, modern F1 cars manage over 6G in a maximum braking state. |
| 06:26 | In the case where we're redesigning the braking package on a vehicle due to performance issues, we may already have g force data. |
| 06:35 | Using this as a reference, it would be safe to say that with improvements, we'll be able to achieve higher Gs. |
| 06:41 | For our example, let's assume we're talking about a vehicle intended for club level circuit based sprint racing with minimal downforce and on a decent semi slick. |
| 06:52 | So a sustained 1.2G should be a reasonable assumption. |
| 06:57 | To be clear, by sustained, we're excluding a spike valve from the data that's only achieved for a very short period of time, caused by an anomaly. |
| 07:06 | We'll also say our car has a wheel base of 2570mm and a centre of gravity height of 450mm. |
| 07:15 | If you can't measure these yourself, our Suspension Tuning and Optimisation course is a great place to learn this skill. |
| 07:22 | Or alternatively, you can probably find suitable values online. |
| 07:27 | If we sub these into our formula, along with the vehicle weight of 1200kg, we can find that for a 1.2G stop, the vertical load on the front axle is increased by about 254kg. |
| 07:40 | Considering the static weight distribution, this now means that 888kg of the vehicle's weight would be over the front axle and 312kg over the rear. |
| 07:53 | Equating to about a 74% front weight bias during our braking event. |
| 07:59 | A quick note on aero before we continue, if we were going to include this in our calculations, it's important to remember that the aero load on each axle doesn't transfer with the weight load we just calculated. |
| 08:12 | So we'd need to add the front and rear aero loads to the values we have now. |
| 08:17 | This does get tricky as the aero load and therefore braking capacity changes with speed. |
| 08:22 | In this case, we'd be designing our system for optimum performance at a certain speed and the driver would need to modulate the brakes as the speed drops. |
| 08:34 | Now remembering that the longitudinal tyre force is equal to the normal force at the tyre which we now have, multiplied by the coefficient of friction between the tyre and the road. |
| 08:44 | As we've also discussed, we want our brake torque capability to be equal to this force multiplied by the tyre radius. |
| 08:53 | The tyre radius will be the same, given we're running the same diameter tyres front and rear. |
| 08:59 | And also assuming that the coefficient of friction is the same front and rear, this 74% weight distribution during our 1.2G stop will be our target brake bias. |
| 09:11 | We still need to calculate the brake torque capacity so we can design our system to achieve this. |
| 09:16 | The next question you might be asking is how do we know the coefficient of friction between the tyres and the road without access to tyre data? The fact is, we've already somewhat defined this by making the assumption that we can achieve a deceleration of 1.2G. |
| 09:34 | The maths actually works out quite nicely in the case of braking because the maximum deceleration is essentially equal to the average coefficient of friction of all the tyres. |
| 09:44 | To clarify what I mean by this, the force on the vehicle during braking is equal to the mass of the vehicle multiplied by the acceleration or rather deceleration. |
| 09:54 | F=MA should should familiar. |
| 09:58 | The force must be equal to the total frictional force at all four tyres which as we know equals the normal force multiplied by the coefficient of friction. |
| 10:08 | The total normal force is also the mass of the vehicle so these cancel each other out and we're left with the acceleration being equal to the coefficient of friction, mu=1.2. |
| 10:20 | The same assumption can't be made for cars with significant aero because the drag produced has a much higher impact on slowing the vehicle. |
| 10:30 | In this case, the tyre's coefficient of friction will be lower than the deceleration value. |
| 10:35 | This aligns with what we discussed earlier about cars with aero being able to decelerate at higher Gs. |
| 10:43 | So let's move ahead with our calculations, starting with the longitudinal forces on the tyres. |
| 10:49 | Reusing our example vehicle during a 1.2G stop, the load on the front axle was 888kg and 312kg on the rear axle. |
| 11:00 | Equating to roughly 74% front weight distribution. |
| 11:04 | Now we can multiply these normal forces by the coefficient of friction of 1.2 and that gives us the longitudinal force of about 1066kg on the front axle or 5228 newtons at each front tyre and 374.4kg on the rear axle or 1836 newtons at each rear tyre. |
| 11:30 | Our car has a loaded tyre radius of 320mm. |
| 11:34 | If we multiply this by the longitudinal force, we'll be aiming for a brake torque capability of about 1673 newton metres on each front brake and 587 newton metres on each rear brake. |
| 11:50 | Which we can see roughly matches our 74% load bias during our braking event. |
| 11:55 | With our brake torque capacity and therefore target brake bias and torque requirements defined we'll round out this module. |
| 12:02 | Lastly I want to quickly note that for vehicles that will run in various conditions, it can be worth repeating these calculations for an estimated maximum G stop in the worst conditions it'd be likely to see to help understand the range of bias that we'd want to have adjustment for. |
| 12:20 | Before we move onto sizing our components, let's recap what we've just covered. |
| 12:25 | If we understand the vehicle's wheel base, centre of gravity height, weight and static weight distribution, we can calculate the longitudinal load transfer and therefore the vertical tyre load for a certain G force braking event. |
| 12:39 | Some assumptions around tyre grip need to be made to avoid over complication but we can then use the tyre load with the coefficient of friction of the tyres to find the longitudinal tyre forces and then with the tyre's radius to find their brake torque capacity. |
