Determine injector
size from known physical parameters
NOTE! Incredibly bold
assumptions – teee heee! HEY, you gotta start somewhere!!
Assumptions:
Highest fuel consumption will occur at peak torque
Peak torque occurs at 3700 rev/min
Peak fuel demand is 18 mi/gal or 6.4 Km/litre
Differential ratio is 3.44:1
4^{th} gear is 1:1
Tyre size is 145x80x10 which is 1060 rev/mi or 660 rev/Km
Maximum duty cycle available for balanced running is 58% of 1 revolution
6.4 
* 
1000 m 
* 

= 
6.4 m 
= 
0.156 ml 




1000 ml 

ml 

m 
3700 
* 
1 wheel rev 
= 
1075 
* 
Km 
= 
1.630 Km 
min 

3.44 

min 

660 

min 
1.630 
* 
1000 m 
= 
1630 m 
min 



min 
0.156 ml 
* 
1630 
= 
255 ml 


min 

min 





So if the injector is running at 58% open time its total
capacity will then be
255 ml 
* 
1 
= 
440 ml 

min 

0.58 

min 

Since those 440 cc are being fed to the engine by 2
injectors, each will have a capacity of 220 cc/min. Using those same parameters
in a spreadsheet we get:
CHECK: The 5.7 L V8 Corvette uses eight 219230 cc/min injectors. That’s an equivalent of a 2.8 L 4 cylinder roughly double our 1.3 L . Since we are using 2 injectors to feed the whole engine instead of 4, that’s about the same as a 1.4 litre. QED