# Exhaust Valve Duration

Pictured above are some valves on the left and the valves in their place in the engine on the right. The valves factor in to the amount of burnt gases that get out of the cylinders.

## What you do:

Choose a duration that gives your engine enough time to let the burnt gases in the cylinder go out.

Factors to consider:

How it all works:

This is a picture of a cam. The lobes on the cam are the egg shaped bumps. These lobes have lift and duration specifications that you dictate. The lift is how high the lobe on the cam will lift the valve. The duration is how long, in crank degrees, the valve will be open.

It is easier to push air out of an engine than it is to draw air in, so the average open exhaust valve area need only be 57.82% of the average open intake valve area. To design a cam and exhaust valve combination with an average open exhaust valve area that is 57.82% of the average open intake valve area, you could estimate it by using an exhaust valve duration or an exhaust valve size or an exhaust valve lift that is 57.82% of the same intake values. However, this will leave you short of the maximum benefit of a well designed cam. You will be missing out on significant power.

So, if you want to be be exact and maximize power (and you can handle the length of the calculations below), it is recommended you use the equations below to make the most of your cam and available valve space.

Exhaust valve duration is determined by the exhaust lobe on the cam. Exhaust valve duration is measured in crank degrees. When we set the duration in crank degrees, we are determining the point on the cam at which the lobe begins pushing the lifter up and where it lets the lifter back down. Since the cam rotates 1 time for every 2 times the crank rotates, an exhaust valve duration of 300 degrees means the exhaust valve is open for 150 degrees of a single revolution of the engine. Exhaust valve duration, exhaust valve diameter and exhaust valve lift factor in together to determine air velocity in the exhaust system. Long duration, large diameter and high lift all cause the engine to breath better at high rpms. Short duration, small diameter and low lift all cause the air in the exhaust system to go faster, but chokes air flow when the air velocity reaches the speed of sound. The speed of sound is 767.58 mph (miles per hour) or 67547.4 fpm (feet per minute).

Since the cam rotates 1 time for every 2 times the crank rotates and cam duration is measured in crank degrees, the HiPerMath equation for exhaust cam degrees is:

exhaust cam degrees = __exhaust valve duration__

2

Next, we calculate the angle from the x-axis to the point at which lift begins. We call this angle "theta".

theta = 90 - __exhaust cam degrees__

2

Next, we calculate the x and y values of the point at which lift begins, using the angle we just calculated:

x = 0.5cos(3.14159/(180 / theta))

y = 0.5sin(3.14159/(180 / theta))

Next, we calculate the b-value for the equation y = ax^{2} + b, which is the equation
for a parabola you have probably seen in your high school math classes. We use a parabola
because it closely represents the shape of a cam lobe:

b = exhaust valve lift + 0.5

The 0.5 in the equation above is the radius of what is called the cam base circle. The cam base circle is the minimum radius the lifter will rest on.

Now that we have values for x, y and b, we can calculate the a-value for the equation
y = ax^{2} + b :

a = -(y - b) / x^{2}

Now that we have the a and b values for the equation, we plug in -0.4, -0.3, -0.2, -0.1, 0, 0.1, 0.2, 0.3, 0.4 for the x values (since the x values for our parabola will range from -0.5 to 0.5, to get values for y:

yn4 = -a(-0.4^{2}) + b

yn3 = -a(-0.3^{2}) + b

yn2 = -a(-0.4^{2}) + b

yn1 = -a(-0.1^{2}) + b

y0 = -a(0.1^{2}) + b

y1 = -a(0.1^{2}) + b

y2 = -a(0.2^{2}) + b

y3 = -a(0.3^{2}) + b

y4 = -a(0.4^{2}) + b

We calculated values for y above because we are going to use them in the distance equations below where we calculate the distance from (0,0) to (x,y) and subtract the radius of the cam base circle:

d1 = &radic ^{2} + yn4^{2})

d2 = &radic ^{2} + yn3^{2})

d3 = &radic ^{2} + yn2^{2})

d4 = &radic ^{2} + yn1^{2})

d5 = &radic ^{2} + y0^{2})

d6 = &radic ^{2} + y1^{2})

d7 = &radic ^{2} + y2^{2})

d8 = &radic ^{2} + y3^{2})

d9 = &radic ^{2} + y4^{2})

Now we calculate the average lift:

average exhaust lift = __ d1 + d2 + d3 + d4 + d5 + d6 + d7 + d8 + d9__

9

Next we calculate average open exhaust valve area:

average open exhaust valve area = average exhaust lift x exhaust valve size x 3.14159

Now we can calculate the air velocity, in fpm (feet per minute) through the exhaust valve, during the time it is open, at any rpm we want:

air velocity through exhaust valve in fpm = __0.327 x number of cylinders x 2 x stroke x rpm x 3.14159 x (bore / 2)__^{2}

average open exhaust valve area x 12

Finally, we can put 67547.4, which is the speed of sound in fpm, in place of "air velocity through exhaust valve in fpm", solve the equation for rpm and find out what the maximum rpm is from that result! The HiPerMath equation is below:

maximum rpm = __ 67547.4 x average open exhaust valve area x 12 __

0.327 x number of cylinders x 2 x stroke x 3.14159 x (bore / 2)^{2}

We just found out how much this cam allows the engine to breath!

Definitions:

fpm = feet per minute

rpm = revolutions per minute