New cars are confusing. With all the computers, sensors, and gadgets, it may seem like there's some sort of magical witchcraft and wizardry taking place under the hood. We're here to show you how modern automotive computer control systems work but today we're going to start with some old tech: The carburetor.
 so almost no new cars use carburetors. Still, it's important to understand how engines got to where they are today. It all began with the good ol' carb. For a lot of you this is review, but if we want a new generation of car enthusiasts to care about cars, it can't hurt to explain how they actually work.
 

To optimize engine performance, engineers want to ensure that enough air is mixed with gasoline so that all of the gas burns during combustion. Such a mixture where all of the fuel is burned is known as a stoichiometric mixture. Maintaining a stoichiometric mixture allows engines to take maximum advantage of gasoline's high energy density (34 mega Joules per liter). If not enough air is provided, the engine will run rich, often resulting in poor fuel economy and black smoke exiting the tailpipe. If there is too much air mixed with the fuel, the engine runs lean, producing less power and more heat. Therefore, engineers must optimize this ratio to gain the most mechanical work per unit mass of fuel. The optimum ratio of air to fuel for a typical combustion engine is about 14.7 pounds of air for every pound of gasoline. The question of how to assure this perfect ratio has been at the forefront of automotive engineering design for decades.
CARBURETORS
p1, ρ1, and v1 are the static pressure, density, and velocity, respectively at point 1. p2, ρ, and v2 are the static pressure, density, and velocity at another location in the flow. We can assume that the density of the fluid is remaining approximately constant, so ρ1 is about the same as ρ2. Let's say that at point 2 downstream, we have a narrowing where the fluid velocity increases. This means v2 is greater than v1. For the left and right sides of the Bernoulli equation to remain equivalent, p1 must be greater than p2. Thus, the high velocity at the narrowing yields low pressure.