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Power assist brake design
My question is simple (then again, maybe not!):
How do you calculate the force on the master cylinder given a specific input force to the power booster? I have some one using the brake worksheet for a power assist, tandem master cylinder setup. I did NOT put the sheet together as it is for this setup. The tandem issue seemed to me to be simple to resolve. Since each piston in the master cylinder sees full input force from the peddle as opposed to a dual setup which sees half (with balance bar centered), simply double the desired peddle force. For instance, if you desire to apply 75 pounds of force on the peddle to stop the car at the maximum rate, change that value to 150 to 'fool' the spread sheet and it will actually calculate the correct values for 75 pounds of peddle force. Now the power assist. I know that the amount of boost will be calculated by the pressure difference between atomospheric pressure and manifold pressure using the power assist diaphram diameter as the piston. However, how does peddle force fit into the equation? Thanks in advance for the help! Rick |
Perhaps I am just making this to hard on myself. I was thinking, power assist, hmmmmmm, then the light bulb went off (stomp on it if this is not correct)!
Is it just as simple as pressure difference on a diaphragm adding hydrualic assistance? For instance if you have one side of a diaphragm exposed to 16"HG (7.86 psi (lb/inch2)) and the other to atmospheric pressure on the other, lets use 14.7psi and a 8" diameter diaphragm is it ((4 * 4) * 3.14) * (14.7 - (14.7 - 7.86)) or 394.8864 pounds of assist? So say you had a system designed to stop at maximum deceleration with 75 pounds of peddle force (6 to 1 peddle ratio) and did nothing but add the booster in the previous example your peddle force required for the same rate of deceleration would be only be 9.33 pounds (I know, not a real world situation)? ((75 * 6) - 394) / 6 = 9.3333333333333333333 Not sure I like the way the amount of assist can vary so greatly with changes in vacuum. Rick |
Sooooo, no one knows anything about power brake boosters? How many people have them on their car?:D
Rick |
Rick, I would like to give some input...but I haven't really ever studied a brake booster system....I don't know how it works...
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It is a mistery! I think my second post sounds logical I am just looking for some confirmation. Then of course the challenge is how do you know the particulars of differenct systems!
Thanks Rick |
Is the booster between the pedal and the master cylinder? Is it kinda made like a plunger?
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That is the type I am refering to. There are all kinds of boosters. Electric, hydrualic, etc. I am interested in the garden variety vacuum assist booster.
Here is a little cutaway drawing: http://auto.howstuffworks.com/power-brake1.htm I think the drawing in the link is misleading though. It has a caption that says "vacuum on both sides of the diaphragm". The darn thing wouldn't work unless there was a pressure difference. You can answer one thing for me. Does my math seem to be correct and is my conversion from vacuum (inHG) to PSI correct? Thanks Rick |
Yep, your conversion is correct. And your math is correct....
However, I don't think it would be a linear pressure application. The way the diaphragm is made, I think it would be almost a natural log function....or half a parabola. Basically what you described up above would be like the surface of a piston in a cylinder. I think this thing has more of a tapered design...right? The harder you push, the more power you're gonna get. You're essentially creating more and more pressure...but not in a linear fashion. The power booster just multiplies the force you place on the master cylinder, right? That's why you would need it in a tandem situation...because it takes more force to compress a single brake master cylinder which is servicing 4 wheels worth of brakes. But I do think the rate at which it does so is not linear...unless the diaphragm and housing are perfectly cylindrical. |
OK....I see how this works now. You do need vacuum on each side of the diaphragm. The diaphragm is encased in a vacuum sealed area....when you apply the brake, the valve opens, allowing atmospheric pressure in...giving you vacuum on one side of the diaphragm and then atmospheric pressure on the other. However, when you release the brake, that valve closes, allowing the vacuum to come to both sides.
If you didn't have the diaphragm encased in a vacuum at all times, your brakes would continually have pressure on them. You need a point there where there is nothing acting upon that diaphgram. |
I think what you'll need to have, Rick, is a compression ratio.....
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Thanks for double checking the math.
Yep, I was looking at it in a linear fashion. Even though I realize it is a diapragm and not a piston. If it is non-linear then I will look for the information to be already published. Some one, some where, knows how much assist these things create. I wonder, with as small of an amount of travel that you typically get at the master cylinder it is non-linear or if it is, is the amount significant. Even with a parabola approach the length to parrellel would not change much considering the amount of movement, the degree of arc we are talking about. Some else just occured to me. These diapragm, at least you would think, would have to be 'rigid'. They are attached at the center and around the circumference to 'fixed' points. As the point in the center moves the distance from the center to outside radius would increase. I seriously doubt the diapragm stretches. With that in mind, while it may be a diapragm I would think that it made in such a fashion that it behaves like a pistion. Just some random thoughts. Of course I could test a car and figure it out that way. Sounds to much like work to me. :D Rick |
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Like I said before, I think the amount of pressure you get varies upon how far the diaphragm is pushed. I think of it just like a piston/cylinder relationship. The further you push it, the further the diaphragm moves, and the more you compress the air inside. Even though the cylinder is not totally sealed like in an engine, I think you would still get a great amount of pressure created by compressing the air...thus moving the linkage to the master cylinder.
Therefore if you had that empirical number on how much you actually compress the air at full pedal depression, you could calculate how much force the air is placing upon the master cylinder. Does this make sense, or am I passing 3rd base headed for the outfield? |
It seems to me that the pedal resistance is dependent on the small spring that controls how far the valve is depressed most of all. Then - you have to figure the spring equivalent of the vacuum effect to get the 'efficiency ratio'.
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This is gonna suck, Rick.
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Thanks Rick |
Let's see
Your math is correct for the mercury to pounds and so forth.
The math for the force is correct as well. As I understand it, braking is done on decelleration and therefore the vacuum is highest. There is a check valve in line with the diaphram ( on my old Ford there was ) so there is assist most of the time. When the pedal is worked often with little or no vacuum, the power assist is depleted. I am not sure, but I think the ratio of 6:1 may be correct. Given this, it would take 10 pounds of force to get an equivalent of 75 pounds. I am sure this is why some vehicles will stop by gently pressing on the pedal. What is your quest? :3DSMILE: |
The quest involves my brake spreadsheet that I am playing with. Some one was attempting to use it with power brakes but it does not have provisions for them. I was hoping to come up with some numbers/formula that I could use in order to help solve the problem.
Thanks Rick |
Okay, short of it is, instead of 75 pounds of pressure to stop the car, all you need is 10.
As I remember it, the cuve is close to a line in the operating regions. When the pedal is pressed, the vaccum assist the rod to the cylinder. When you let off, there was a spring that retracted the diaphram. I think you can address this as an asymtotic ( you really have to love my spelling ) equation with a long curve. Something like e ** x. "e to the x" I get the impression you are fixing the equation in the sheet. if this is correct, Why? |
As I understood it.
As I was told by one of the guys in chassies, there are several factor involved in the brake system design.
Weight of the car, friction cooefficient, size of the intended tires, center of mass and so on. Once all that is done, the master cylinder and ratio of front to rear is mixed. Then the amount of pressure to activate the system is measured. Then the guys compares that against the PDD or CTS. The booster is selected to provide reasonable braking with reasonable foot pressure. Remember, This is very old info. For the most part, components in a car or truck are selected by : 1. Price. 2. Across the board use. 3. PDD or CTS. 4. Functionallity. The whole reason the brake booster was design the way it was because it was cheap and they, car makers, needed to improve their rotten braking systems. The vacuum of the engine was untapped as a "power source". In the 70s you had vacuum HVAC controls. brakes and interior adjusters. I undertand they even had vacuum assist windows ( wonder if Bill Gates knows this ) :LOL: :3DSMILE: |
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