since i just went though this, i thought i should write down what i learned before i forgot it. hopefully it will help some people ... please accept my apologies beforehand for how long and rambling it is ... so here goes ... the injector timing table is based on crank degrees, and specifies the degree at which the injector pulse should end. so at what we is a way to figure out a process that gives us, for a given rpm and load, what cam degree the injector pulse should end. note that crank degrees go from 0-720 instead of from 0-360 because the crank rotates 2 times for each revolution of the cam. when there is little or no overlap between the closing of the exhaust valve and the opening of the intake valve, it may be desirable to actually start injecting the fuel before the intake valve opens. in this case, the fuel is squirted onto the back of the closed intake valve, because that may help atomize the fuel. but if there is alot of valve overlap, then that approach is not as desirable, because when the intake valve opens, some of the fuel that was waiting to go into the cylinder will get mixed in with the exhaust (and ejected from the cylinder) and therefore not be burned, potentially resulting in a lean condition. so a factor in deciding how early to start the injectors may be how much overlap there is in your cam. here are my thoughts on a process to determine what values to put in the injector timing table ... before we can pick an ending degree, we need a way to determine how many degrees the injector needs to provide the right amount of fuel. in order to do that, we need a way to figure out how much fuel is needed. once we know that, we can determine how many milliseconds the injector needs to be open to provide the fuel. then need to be able to figure out, for a given rpm, how many degrees the camshaft will turn during the time the injector is open. then, knowing the degrees at which the valves open and close, we can finally pick reasonable ending degrees for the injector timing. for a certain rpm and load, we can estimate what the injector pulsewidth would be. for example, suppose we had a 410 with 42# injectors spinning at 3000 rpm under 50% load. what would the estimated pulsewidth be? one cylinder has a maximum volume of 410/8, or 51.25 ci. since the volumetric efficiency of the engine is not 100%, for our example, let's use .85 as the VE factor, which means that a given cylinder will end up with a maximum of 51.25*.85 = 43.5625 ci of air. since our example is at 50% load, that would be 43.5625 *.50 = 21.78125 ci. this is the estimated amount of air (in cubic inches) each cylinder should receive per cycle. there is a conversion factor that says one cubic foot of air contains .075 lbs of air. so in our example, we have 21.78125 / (12*12*12) = 0.01260489 cubic feet of of air. so we have 0.01260489 * .075 = 0.00094536675 lbs of air now that we know we have 0.00094536675 lbs of air in the cylinder, we can determine how much fuel we need ... using the standard air/fuel ratio of 14.67, we would need 0.00094536675/14.64 = .0000645742316 lbs of fuel. since injectors are given in lbs per hour, and pulsewidths are in milliseconds, we need to convert to lbs per ms. so for 42# injectors, we have 42/(60 minutes*60 seconds*1000 milliseconds) = 0.0000116666667 lbs of fuel per ms. so to get .0000644421779 lbs of fuel, we would open the injectors for a pulswidth of .0000645742316/0.0000116666667 = 5.53493412 ms. but since there is a small delay (about .05 ms) before the injectors actually start squirting the fuel, lets add .05 to the pw, getting about 5.58 for the desired injector pulsewidth. notice that if we figure out the pw for the case where load is 100% (WOT), then we can just multiply that by the load later to simplify things. let's do that here: max air in lbs: (51.25*.85*.075)/(12*12*12) = 0.00189073351 lbs air fuel needed for max air above: 0.00189073351 / 14.67 = 0.000128884357 lbs fuel fuel in lbs per ms: (42/360000) = 0.0000116666667 ms needed to deliver the above fuel: (0.000128884357 / 0.0000116666667) = 11.05 ms note that we will need to add the .05 ms delay to the pulsewidth after factoring in the load ... so 11.09 ms is needed to deliver the most fuel the injecter should have to deliver. of course, this will be slightly higher if we want the afr to be lower. like when at WOT for example. we will multiply this to the load later ... ok, we are halfway there... now we get to figure out how many degrees the crank will turn during that 5.58 ms. that depends on how fast the engine is turning. in our example, the engine is turning 3000 rpm. since there are 360 degrees in a revolution, 60 seconds per minute and 1000 ms per second, then we have the crank turning at (360*3000) / (60*1000) = 18 degrees per ms. so in 5.58 ms, the crank turns 18 * 5.58 = 100.44 degrees. notice that the only variable above is the rpm, so for a given rpm we can do this to get the degrees per ms: 360*(rpm)/(60*1000) = 360*rpm/60000 = rpm*(360/60000) = rpm*0.006 we'll use this later too ... so let's say that our cam has an intake lobe center of 110, a lsa of 112, a duration of 230 at .050" and 290 total duration. so how does that translate into crank degrees? here's how: by definition, crank degrees 0-360 refer to the first revolution of the crank, and degrees 360-720 refer to the second revolution of the crank. intake events are offsets from top dead center (TDC) or bottom dead center (BDC)of the cylinder during the phase in question. for the intake valve events, TDC is at degree 360. since the intake lobe has a center of 110, that means the valve is open its highest at 360+110=470 degrees. a duration of 290 means that the valve opens at 470-(290/2) = 325 and closes at 470+(290/2) = 615 degrees. similarly, the valve is at .050" or higher from degree (470)+/-(230/2), or 355 and 585 degrees. a lsa of 112 means that the exhaust lobe center is 112*2 degrees before the intake lobe center. so with a lsa of 112, our exhaust valve is fully open (the exhaust lobe center) at 470-(112*2) = 246 degrees. doing the same calcs as above, we have open/close at 101 and 391, and at .050" the open/close events are 131 and 361 degrees. now notice that the exhaust closes at 391, and the intake opens at 325. so both valves are open for 391-325 = 66 degrees. this is the total overlap. similarly, at .050" there is an overlap of 6 degrees. in this case, i would probably want the fuel to start injecting when the exhaust valve closes, or at about degree 390. now we are almost there ... for a given load and rpm, we can figure out how much ms of injector we need, and how many degrees that ms needs. now all we have to do is tie that to the cam events for our cam and we can fill in the table with values that should be fairly reasonable. to finish off our example, with the 410ci engine and 42# injectors, 50% load at 3000 rpm with our cam as above, the calculation for that cell would be ... the needed injector pw would be: 11.05 * .50 + .05 = 5.595 ms the cam rotation for 5.595 ms at 3000 rpm would be: 3000 *.006 * 5.595 = 101 degrees (rounded) if the pw starts when the exhaust valve closes ... the injector fires from degree 390 thru 491 (390+101) so 491 should be a fairly reasonable value for that cell.