Brake System Design and Optimization: Brake System Hydraulics
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Brake System Hydraulics
06.42
| 00:00 | - For both road cars and motorsport applications, hydraulic pressure is used in almost all cases to transfer a force from one braking component to another. |
| 00:09 | And there's a number of advantages to using hydraulics rather than other systems like push or pull rods and cables. |
| 00:17 | These include simple force multiplication, flexibility in routing and packaging, instantaneous transfer of force with essentially zero losses and long service life if maintained appropriately. |
| 00:31 | In this module, we're going to take a look at how hydraulic brake systems work and what we need to be aware of when upgrading our brakes or designing a new system from scratch. |
| 00:42 | The most important principle to understand in hydraulics is the relationship between pressure, area and force. |
| 00:50 | By exploiting the relationship between these parameters, we can simply change the ratio between the input and output forces. |
| 00:58 | The relationship looks like this. |
| 01:01 | Pressure equals force divided by area. |
| 01:03 | The trick to this simple equation is that if we know any two of these three parameters then we can easily calculate for the third. |
| 01:11 | This is simply done by rearranging this equation for the parameter of interest as we see here. |
| 01:18 | Let's look at some simplified examples to understand this principle. |
| 01:22 | Take the case of a simple cylinder filled with water that has a piston acting on it. |
| 01:28 | There's a static force applied to the top of that piston, let's say this force is equal to 10 newtons. |
| 01:34 | To keep the math easy, let's also say that the surface area of the piston is 1m² . |
| 01:40 | Now that we have the applied force in the area of the piston, we can solve for the pressure in the fluid due to the applied force. |
| 01:48 | Subbing these numbers into the equation, we get 10 pascals which is the standard SI unit for pressure. |
| 01:55 | Now let's extend this example a little further by taking our original cylinder and connecting it with a small hose to another cylinder of the exact same dimensions. |
| 02:06 | The area of the piston in the second cylinder is identical to the first. |
| 02:11 | An important principle in hydraulics is that the pressure in the fluid due to the force acting on the piston will be equal throughout the entire system regardless of where we check it. |
| 02:20 | This means if we measure it in the hose, the original cylinder or the new cylinder, we're going to see the same pressure. |
| 02:29 | At this point it's worth introducing some terms that'll be useful as we move throughout the course. |
| 02:34 | The original cylinder, the one that we're applying the force to is generally referred to as the master cylinder. |
| 02:42 | The new cylinder is referred to as the slave cylinder. |
| 02:45 | This is the cylinder that does the useful work for us. |
| 02:48 | If we want to know the force exerted on the piston of the slave cylinder, we can use the same equation as before but rearranged. |
| 02:57 | You can probably guess what it's going to be but let's plug in the numbers and check to see how it works. |
| 03:03 | We know that the pressure in the system is 10 pascals and we know that the area of the slave cylinder is 1 metre squared. |
| 03:10 | Using one of the rearrangements of the original equation, we can see that the force the slave cylinder will exert is 10 newtons. |
| 03:18 | We get the same force out of he slave cylinder as we input into the master cylinder which makes sense. |
| 03:25 | Obviously we're looking at an idealised system here and we're ignoring any friction. |
| 03:30 | However for the calculations we'll be doing in this course, this is still a safe simplification. |
| 03:36 | Now let's make things a little more interesting. |
| 03:38 | We'll keep the master cylinder size the same but we'll double the piston area of the slave cylinder. |
| 03:44 | Looking at the numbers again, we have the same input force which obviously gives us the same fluid pressure. |
| 03:51 | However, now when we solve for the output force at the slave cylinder, we've doubled the output force to 20 newtons. |
| 03:58 | At first glance it might seem like we've got something for nothing here. |
| 04:02 | Simply by changing the slave cylinder bore size, we've got an extra output force for free without changing the input force. |
| 04:10 | The way we pay for this though is that while the output force has increased, for a given displacement of the master cylinder piston, the amount of displacement of the slave cylinder has decreased. |
| 04:22 | Halved in this case. |
| 04:24 | Teh opposite is equally true. |
| 04:26 | If we had instead halved the area of the original slave cylinder piston, the output force would only be 5 newtons but the displacement of the slave cylinder would be double that of the master cylinder. |
| 04:37 | If we now connect a second slave cylinder into the system, we can see that the same math applies. |
| 04:43 | Let's say both slave cylinders have the same piston area of 2 metres squared. |
| 04:49 | For our same input force at the master cylinder, we have the same hydraulic pressure throughout the system of 10 pascals. |
| 04:57 | This results in each of the slave cylinders exerting the same force of 20 newtons. |
| 05:02 | This is an important point to remember, adding more slave cylinders into the system does not reduce the force each of them can produce. |
| 05:11 | This is simply a function of the fluid pressure and the piston area. |
| 05:15 | What it does mean however is that for every extra slave cylinder we add into the system, the resulting displacement of each slave cylinder will reduce for a given amount of master cylinder displacement. |
| 05:28 | Hopefully it should be clear from these examples how we can easily achieve different output forces for a given input force by changing the bore sizes of the master and slave cylinder. |
| 05:40 | This is one of the main advantages of using hydraulics for a system like brakes. |
| 05:45 | By giving us the simple means of changing the force ratios. |
| 05:49 | Before we move on, we need to mention another important assumption that we'll use throughout the course. |
| 05:55 | Which is that the fluid we use in the hydraulics is treated as if it's incompressible. |
| 06:00 | This simply means the fluid volume does not change when pressure is applied to it. |
| 06:06 | For our purposes, and assuming the brake system is working correctly, this is a reasonable assumption although we will be discussing some cases later in the course in which this doesn't hold true. |
| 06:19 | In summary, the property of hydraulics make them suitable for use in braking systems. |
| 06:23 | There is a simple equation that governs the relationship between the pressure, force and piston area. |
| 06:30 | By varying the piston area of the different components in the hydraulic system, we can change the input and output force relationship between components. |
