Intake diameter
- Thread starter Gavin
- Start date
I don't believe they're round, so a direct diameter may not be possible. An oval/ellipse is more like it and should be able to get a very close estimate by that means.
flat_black may be a good source for that (or anyone else that has the head/intake mani off).
Doing some volumetric efficency calcs?
flat_black may be a good source for that (or anyone else that has the head/intake mani off).
Doing some volumetric efficency calcs?
twilightprotege
Member
- :
- 2001 323 Astina SP20 (P5)
yes they are oval in shape
Yup,
I'm looking at starting up a small NA project. Nothing terribly drastic, Im just looking to improve my throttle response.
I assumed that the intake ports on the head would be more ellipsoidal than circular. I thought they might taper out to a circular cross-section over the intake runner length though. Either way, if anyone can enlighten me with some intake runner/port area information Id appreciate it.
flat_black
Member
Well, I'm ASSUMING the intake port is 35mm by 50mm... I never measured 100 percent! I totally spaced that! Ahh, well.
At any rate, assuming it's 35 by 50 oval, we'd take the 35, and use that as the diameter, giving us a 17.5mm radius.
17.5mm in radius gives us the critical info for the pi*R^2 equation, leading us to the following:
pi*17.5 = 54.978
54.978^2 = 3022.58... = Area of the radiused ends.
Now we calculate for the area between the radius.
50 - 35 = 15 = The width of the rectangle between both radii. Height is eqivical to the diameter, 35, so:
35 x 15 = 525
And in conclusion:
3022.58 (area of the radius) + 525 (area of the rectangle) = 3547.58 (total ovoid area)
But those are just based on guesses, as far as sizes. I would dare say I erred on the large size.
Oh, and the results would be expressed in mm^2. (Millimeters squared)
At any rate, assuming it's 35 by 50 oval, we'd take the 35, and use that as the diameter, giving us a 17.5mm radius.
17.5mm in radius gives us the critical info for the pi*R^2 equation, leading us to the following:
pi*17.5 = 54.978
54.978^2 = 3022.58... = Area of the radiused ends.
Now we calculate for the area between the radius.
50 - 35 = 15 = The width of the rectangle between both radii. Height is eqivical to the diameter, 35, so:
35 x 15 = 525
And in conclusion:
3022.58 (area of the radius) + 525 (area of the rectangle) = 3547.58 (total ovoid area)
But those are just based on guesses, as far as sizes. I would dare say I erred on the large size.
Oh, and the results would be expressed in mm^2. (Millimeters squared)
Flat_black, I just noticed a problem with your math.
Area of a circle is pi*r^2, not (pi*r)^2
So the area of the radiused ends would be:
3.14 * 17.5^2
= 3.14 * 306.25
= 961.625mm^2.
Total area would then be closer to 961.625 + 525 = 1486.625mm^2
Im not trying to piss on your help or anything, just noticed that small error.
Area of a circle is pi*r^2, not (pi*r)^2
So the area of the radiused ends would be:
3.14 * 17.5^2
= 3.14 * 306.25
= 961.625mm^2.
Total area would then be closer to 961.625 + 525 = 1486.625mm^2
Im not trying to piss on your help or anything, just noticed that small error.
flat_black
Member
I was wondering why that seemed so high to me! Hehee... I was a bit tired when I typed that, so... 
