# Cam lobe question

1. Oct 4, 2006

### moo

Hey guys,

Let's say we have two cams, and everything is identical (rpms, valves, springs, etc.), except the angle of lift on the cam.

For simplicity, let's assume the lobe lifting edges are straight, and we want to achieve a lift of .75" (19 mm) with both cams.

Cam #1 has a 60 degree angle of lift.
Cam #2 has a 45 degree angle of lift.

Question:
I understand that cam #2 would open the valve slower, but do both cams require the same amount of total power to complete one lift cycle?

Thanks, moo
__________________
moo (moo') adj. Of no practical importance; irrelevant, such as a moo point (i.e. a cow's opinion).

2. Oct 4, 2006

### marcusl

Are you saying that the ramp on either side of the high lift point has a 45 (or 60) degree angle? If so, then the net energy in opening and closing the valve is zero, because energy put into compressing the spring is regained when the valve is closed. (We are neglecting friction--assume good lubrication and, say, a roller lifter). The power is constant plus and minus since the ramps are straight, so it might look like

----

-------- -----------> time

----

for the 60 degree lobe. The power will differ with the two cams because you are putting energy in slowly with the lower ramp angle, and getting it back more slowly too, but the area under each part must be the same. (Sorry I can't draw it to scale here )

------
------ ------> time
------

3. Oct 4, 2006

### marcusl

Oh no. My diagrams didn't work at all! Imagine a single square-wave up and down, and another one that is longer but not as high.

4. Oct 4, 2006

### moo

Hey Marcus,

Thanks. Actually I was more concerned with the "lift" part of the cycle (valve closed to valve open), since that's where external power is required. And I wasn't counting friction either.

But I see the answer would be the same. If we powered these cams with two identical electric motors and the same gear reduction, then it would take the same amount of total watts/power to open the valve with either cam.

Lol, the square wave analogy made the light bulb come back on.

Btw, you can use the "code" tags to do ASCII art, charts etc., just draw it in notepad or something with a fixed width font then paste it.
Code (Text):

__
|  |
__|  |   __
|  |
|__|

_____
__|     |      __
|_____|

Thanks, moo
__________________
moo (moo') adj. Of no practical importance; irrelevant, such as a moo point (i.e. a cow's opinion).

Last edited: Oct 4, 2006
5. Oct 4, 2006

### marcusl

Wait, that's not quite right. The power is the height or amplitude of your square wave and differs for the two cams. You need to expend more horsepower to open the steep profile, albeit for a shorter length of time. It is the energy (area under the curve) that is the same; it has to be since you are compressing the same spring the same amount each time.

BTW, there's one small detail I didn't want to get into right away. If your cam profile is triangular like I described, then the power curve isn't actually square but has a curve to it. This doesn't affect the conclusions above, however, and the square wave is easier to think about.

6. Oct 4, 2006

### moo

Yep I understand this point, that's why I mentioned total power required to open the valve.

Thanks, moo
__________________
moo (moo') adj. Of no practical importance; irrelevant, such as a moo point (i.e. a cow's opinion).

7. Oct 5, 2006

### marcusl

Moo, I think you have the right idea, but I'm not sure because you are misusing some terminology. What you are calling "total power" is the energy expended, also called work. The physics definition of work is the time integral of power, which is the area under the curve in your case.
Conversely, power is work per unit time,
P(t) = dW/dt.
Check out Mechanical Power on Wikipedia
http://en.wikipedia.org/wiki/Power_(physics)

Power is an instantaneous quantity. You need more power to use the cam that lifts the spring more quickly, and would also need more power to spin the motor at higher RPM.

Are we on the same page?

8. Oct 5, 2006

### moo

Lol ok, yeah I think we're on the same page - just different books.

Total simply means "a whole quantity; an entirety."

From your link: "The units of power are units of energy divided by time."

So it seemed logical to me that "total power" is the sum of the units of power in question (such as those required to open a valve). My electric company certainly has no problem figuring total watts (kW actually), which is how power is measured.

But then again, if I were an engineer I prolly wouldn't have posted this question in the first place.

Thanks again, moo
__________________
moo (moo') adj. Of no practical importance; irrelevant, such as a moo point (i.e. a cow's opinion).

Last edited: Oct 5, 2006
9. Apr 10, 2008

### moloni

You are forgetting that the force to overcome the inertia of the valve increases with the square of the speed.
At certain RPMs way below the engine's redline the force needed to overcome the spring force is less than the force needed to overcome the inertia of the valve.

So the answer to your question is that the energy consumption of an, as you call it, high angle camshaft is greater than that for a smaller angle camshaft.

10. Apr 10, 2008

### NateTG

Actually, the electric companies charge by Killowatt hours. (Note the multiplication by time there.)

It makes a more sense to talk about 'average power' than about 'total power'. It's pretty clear for example that a F1 car that gets from 0 to 60 in 2.5 seconds has more power than a VW bug of the same weight that takes 12 seconds to do the same, but if the F1 car spends the balance of those 12 seconds cruising at 60, then the cars have produced the same average power over 12 seconds.

P.S. Moloni is confused or incorrect. Ignore him.