Need mathematical confirmation, please.

[Anonymous]

A=3.14xr squared.

So for venturi one (.513" radius):

A=.885"

And venturi two (.813" radius):

A=2.076"

For a total of 2.961" or approx. a 2" dia. hole, right?



-Chris

 

[Anonymous]

For a total of 2.961" or approx. a 2" dia. hole, right?
-Chris

Chris,
2.961 is alot closer to 3" than 2"

 

[Anonymous]

No, I mean the area for a 2" dia hole is approx. 3.14".

So, the total area need for the two venturiis would be 2.961" or a 2" dia. hole, right?



-Chris

 

[Anonymous]

A=3.14xr squared.
So for venturi one (.513" radius):
A=.885"
And venturi two (.813" radius):
A=2.076"
For a total of 2.961" or approx. a 2" dia. hole, right?
-Chris

Chris,

The equation for the area of a circle is (pi x r^2), or:

3.14159 x .513 x .513 = 0.827 square inches for the smaller venturi, and

3.14159 x .813 x .813 = 2.076 square inches for the larger.

Together they total 2.9 square inches. That is very close to the area of a 2" diameter circle, which is 3.14159 square inches.

Hope that helps!

 

[MLINE]

2" hole is correct.

 

[Anonymous]

Yes, thats' what I was asking.

Thanks.



-Chris

 

[XFlow_Fairlane]

no
you are wrong.

I assume this is for some sort of flow calculation and a 2" hole will flow much more than two smaller holes due to boundary layer thickness. but the math as far as area is concerned is right....just migh tnot be applying it the correct way

 

[Anonymous]

No, I know a 2" dia. hole will flow more than two smaller dia. holes combined. I was just asking for confirmation on the area calculations.

I'm just trying to think of some practical ideas for adapting a two venturi carburator to the "log". I'm not much of a machinist and I would like to keep it simple - simply because of that fact.

Not sure if you are familiar with Asian carburators but they have large throttle bores that funnel down into two venturis. This is the same basic principle that I am thinking of adapting - but in reverse. 2-1.



-Chris