Maximum Injection Time ( ms)
08-07-01, 06:34 PM
#1
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Maximum Injection Time ( ms)
Regarding the formula for calculating the maximun injection time for a rotor engine if I want to rev up to 9000 rpm do I multiple the rpm by 4 or by 2
(120,000 x 2)/ (9000 x ?) =ms
Maybe someone knos about this
(120,000 x 2)/ (9000 x ?) =ms
Maybe someone knos about this
08-16-01, 03:01 AM
#2
Banned. I got OWNED!!!
This is the formula I use
opentime ms (max available) = 1/(RPM/60)X2X1000
This will give you the maximum available time to inject fuel for any rpm you wish, with my autronic system I can reliably run to 95% duty cycle. eg 8500rpm I have 14.117647 ms (from the formula) to get my fuel in...and I have data logged an injector opening of 13.42 ms.
Only use this method if your data logging is accurate, otherwise do not run over 80% duty cycle as you will run into trouble !
hope this helps you out.
opentime ms (max available) = 1/(RPM/60)X2X1000
This will give you the maximum available time to inject fuel for any rpm you wish, with my autronic system I can reliably run to 95% duty cycle. eg 8500rpm I have 14.117647 ms (from the formula) to get my fuel in...and I have data logged an injector opening of 13.42 ms.
Only use this method if your data logging is accurate, otherwise do not run over 80% duty cycle as you will run into trouble !
hope this helps you out.
08-18-01, 02:49 AM
#4
Banned. I got OWNED!!!
This is the formula....
You gotta remember that you are feeding two chambers per rotor per unit of time not one, this causes a lot of confusion between many people (even so called experts!).
You have to compare an apple to an apple so to speak, no good comparing an apple to a water melon!
It is a bit like the rotary engine displacement issue, sure it displaces 1.3 lt in 13B form but it does double the amount of work in the given unit of time that we use to measure engine power/efficiency even though it is a four stroke engine!
I like to call it a 4 cycle 2 stroke, cause this is realy what it is.
Use my formula, I have proven it to work
You gotta remember that you are feeding two chambers per rotor per unit of time not one, this causes a lot of confusion between many people (even so called experts!).
You have to compare an apple to an apple so to speak, no good comparing an apple to a water melon!
It is a bit like the rotary engine displacement issue, sure it displaces 1.3 lt in 13B form but it does double the amount of work in the given unit of time that we use to measure engine power/efficiency even though it is a four stroke engine!
I like to call it a 4 cycle 2 stroke, cause this is realy what it is.
Use my formula, I have proven it to work
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