Encore I was done with this deal but you bring up one question that i think is the real crux of the matter for you, and perhaps I can help clarify it.
You asked something about if a coyote covers so much distance in .12 seconds that you still need to lead him 4 feet no matter what direction he is traveling.
This is the real question and I can see where it could be tricky ,so perhaps I can help clear it up for you.
No you don't hold 4 feet no matter what direction he is traveling.
For several reasons.
#1 is that we do or should know how fast our bullet is traveling , we don't know for a fact how fast any given critter is moving.
That is just one area where " feel " has to come into play.
#2 Distance to said critter, is he 75 yards or 125? now combined with the uncertain speed we have an uncertain distance.
#3 the main issue on your figures that must be taken into account, how much to lead?
Well we have a few constants here that we can take into consideration regarding the "math " of the lead.
One is that we do know that if a critter is at a specific distance moving a specific speed in a 90 degree angle across in front of you then yes you should be able to lead him a specific distance to achieve a hit.
However here is where it gets tricky.
Lets assume that you have a varmint traveling at oh lets say 32 mph do we lead him 4 feet? well now this guy is running straight at you! or straight away.
How much do we lead him? why none at all.
So now we have this same critter doing the same thing only at a 45 degree angle relative to our position, how much do we lead? well he is still for the sake of argument traveling at 32 mph, do we lead him 4 feet? no we don't.
Reason is that his speed may be 32 mph but his speed in relation to a right angle has changed because for every 10 feet he travels foward his crossing distance would only be half of that if he were leaving you at a 45 degree angle.
So if you assumed he is traveling at 32 mph and at that speed at approx his distance it would require a 4 foot hold if he were crossing at a 90, at a 45degree departure the hold would be 2 feet in front.
So I guess yes there is some math that can be done, but there is nothign like seeing dust fly behind one to bring it all home.
