Hitchcocks Forum

Could anyone with a degree in mathematics answer me this...


If static ignition timing is .8mm before TDC (or 1/32") how does that translate into degrees? I'm referring to a 2001 500 Bullet Classic. The reason I want to know is that I don't have a dial guage but do have an old fashioned timing disc.


Cheers


Tony R. 

Riggers. I think that equates to 20' degrees


Hemant 

Riggers, its 10degrees.


ChrisD

Hi guys------------best timing information for practical running purposes is the fully advanced setting--------after all, your engine will be running at this ign. setting more than the idling one. Anyway Hemant & Chris D the question was How does that translate into degrees ie How is it done ?? ----you already have been given  the dimentions of all 3 sides of the triangle formed so it's just a matter of simple trigonometry. ( At top-dead centre the vertical line is Crank throw plus con-rod length ) Move the piston down by 0.8mm then the length of the vertical now becomes (Crank plus con-rod) - 0.8mm, whilst the other two sides remain the same in length. I can't find my Log books but my instinct says that 10 degrees is probally the better figure. Can anyone work this out at all ??

 Tony,


  Unless your 2001 Bullet is something really special, high comp, big cams and a race proven engine, it makes no difference. Low comp, long stroke, slow revs it'l fire somewhere near enough to make the piston go down every time it fires give or take a bit

Alan - Thanks for the geometry lesson! There was a time, long, long ago when I might have been able to work it out. Sadly now I'm afraid it's all been lost in the annuls of time(ing).


Norm - The reason I am going to check the timing is because when starting the bike from cold it invariably spits back several times before firing up. I'm told that this is not good news as something somewhere does no like running backwards. And that, due to this, one day my bike will implode and leave me looking over a garage floor of devastation.


Tony R.

------------"lost in the anus of time ????" Oh, sorry--typo again. Incidently, take 1mm to be 40 thou (0.040") then 0.8mm = 0.8 x 40 = 0.032" OR 1/32"approx. You can do that by ruler and eye-ball at the kerb-side if needs be.----- Have a flexible hack-saw blade with 16 tpi to hand in your tool box. 2 teeth = 0.032", Simples.   As Norm says, you really don't need to be so accurate with the static timing. It's such a low-tune, "cooking" engine that, on occasion I've used just a "nudge" backwards to do a quikie setting. ( OOhhh ! You can't beat a good quickie, that's what I always say ). OOOpppss----sorry Mr mediator. Handy tip-----on an engine with a starter ring gear, count the No. of teeth, divide by 360 and there's your degrees per tooth.

CORRECTION-----------------a blade with 32 tpi then one tooth = 0.032"  DOH !!!!

The technical notes section should give you all the information you need. When I replaced the points to Boyer I timed my 2000 500 with just those notes, a strobe timing lamp and a school protractor.


 It starts and runs beautifully and has never been a problem since.

0.8 mm equates to 10 degrees btdc.


Full advance of 8.5 mm equates to 32 degrees btdc.


As a matter of interest, if you set spark at exactly tdc, you then need to move the points back plate one eigth of an inch measured on it's periphery to get near as dammit 10 degrees of advance;(if you're doing the maths then remember it is geared down 2:1 from crank!) ...... Well, I think that's right, and it certainly seems to work!!