ok, i see what you mean now...and i have a question: how do you get the maximum CE voltage out of a transistor (in respect to BE connection)?
and to your question: a relay, as i have observed, needs X amount of volts to get it to latch, turn on. that is the hardest part of the prosses, from there on it requires less voltage to "hold" itself on, and the very minimum of that voltage is the "drop out" voltage, or the voltage at which the electro magnet can't hold the latch and thus the relay returns to its original state. this is very similar to you pushing a brick up a hill, it requires more force to push the brick than it does to keep it from moving back down the hill, and if you apply even less it will start to move backwards.
i was trying to take advantage of this in my circuit, by using the 9v to switch on the relay i get it to latch (no crap) then i can use a small 1.5v button battery to hold that latch, 1.5v>.25v, and thus freeing the 9v from relay duty and making it short over the common-NO pins and through the igniter. for this to work i would need a <=5v DPDT relay and a way to make the timer pulse once, or else on every pulse the 9v would try to turn the relay on and take voltage away from the igniter. problems: i don't have a <=5v DPDT relay and don't have the money, or will, to get one (only place i know, 6.25$ shipping, .75$ part...worth it? no...) and radio shack only has a 12v DPDT. every step of the way i run into problems, and most relate to power sources and them being too big to fit in the space or too weak and expensive to do anything for very long... i need to do a lot more testing. and too think, if that 9v could hold the relay on AND ignite the igniter this would be over in a second. and work twice as good.