update on tbi setup

[Bruno!]

On Fri, 8 Aug 1997, James Boughton wrote:

[snipped bit about intake]

> Do you intend to have any ability to adjust length for development?  If not,
> I would suggest a length shorter that what you think is correct.  Power can
> fall off very dramatically once the runner becomes too long, but tends to
> fall off much more gradually if the runner is too short.

I quickly went to the garage and measured the length of the runner. The 
manifold part is 155 mm average, and the throttles with the trumpets will 
be about an extra 100 mm. I am going to try use the trumpets that are on 
the carbs right now, otherwise, i could whip up some, of different 
lengths if necessary. 

> 
> >vol of air for each cyl = (500cc/2) * (6600/60) = 27.5 l/s = 0.0275 m^3/min
> >area of tb = (area of tb id) - (area of shaft) = 9.366 x 10^-4 m^2
> 
> In actuality, the average area of the intake system should be used if you are
> going to use the calculations in the SAE papers by Engleman and Eberhard.

If i assumed the area of the tb was the same as the runner due to the
reduced area of brass at WOT, then the cross sectional area would be
approximately 1000 mm^2, which put into the formula below would give 27.5
m/s. 

> >air speed = volume / area = 0.0275 / (9.366 x 10^-4) = 29.36 m/s
> 
> Very high!

> The figure used is 60 m/s over the intake cycle which is 4 times the
figure you > are using because the intake event is only one of the 4
events of the 4-stroke > process.  This would put your throttle bodies at
about twice what I would suggest. > It is really the throttle blade that
causes most of the problem, but what can you do? 

[stuff on port theory snipped]

Before the engine was screwed together my brother and i tidied up the 
intake and exhaust ports. the intake was port-matched, and left rough, 
until just near the valve. all casting marks and irregular spots in the 
ports were removed. Nothing drastic.

If you take the amount of air sucked into one cyl (500 cc) and put this into 
equation, this gives 55 l/s air flow, and over the area, gives about 60 m/s. 
This is the speed over the intake cycle. The division by 2 in the equation 
earlier i understood was from the intake event being half a revolution.

Comments most appreciated. Thaks very much.

		 Bruno. (b.marzano at student.canberra.edu.au)

                On the internet, no-one can hear you scream.