Hard pedal

I heard my name; math coming. But really it's a non-issue.
Since I don't know your system I'll go with the Mopar KH 4piston set-ups
Part unos;
Lets say; seal retraction is .0025, and bearing run out allows .0015 knockback on each side, and pedal ratio is 7:1(a guess).
Bore size on these is 1.636. The area calculates to 2.102/ cavity. So;
[ 2.102 area x (.0025stroke plus.0015knockback)] x 4pistons =.0336 in cubic
That x 16.387 =.55 cc
To that you add the otherside, also ..55cc, totals 1.102cc or.067 cubic inch
Now lets go to the M/C. It is listed as 1.03125. This has an area of .8352 sq.inch
For it to displace.067 in cubic,it would have to travel .067 /.8352 =.081 inch travel. That is just about 5/64 inch of piston travel,inside the M/C. that's a fat sixteenth.
Add the C-port allowance,(during which no braking takes place), maybe .125, that totals;.206 inch
With a 7:1 pedal ratio that is 1.44 inch pedal travel. Add a bit for hose expansion,system flex, and pedal bushing slop; and, I get less than 2 inchs pedal travel, for sure.
This KH system works very well with a 15/16 M/C.(I know cuz I'm very happy with mine)
15/16 is .6903 area. So then .067/.69 =.097 inch travel or about 20% more. .097 is a tad over 6/64. Go get your tape measure to get an idea of what that looks like.Again adding .125 for the C-port I get .222 inch
With the same 7:1 pedal ratio this is now 1.55 inch pedal travel.Plus allowance for expansion and flex; and still similar pedal travel,say a tad over 2 inches.
-Now, to get a sense of this; the last time I bench-bled a m/c, I'm gonna guess the piston stroke was over a half inch: I'll guess .600.If you subtract, IDK, say 1/8 to get past the C-port. You are still left with .475 inch. So the math on this says the M/C has nearly 5 times as much stroke as it requires. That would be nearly 400% additional travel.
What this shows is that even with generous seal-retraction, and sloppy loose bearings(total .003), almost any Mopar M/C can displace enough fluid. And the reservoir size has nothing to do with the equations; only time to top up.
To this you can add some rotor run-out;although I can't imagine why you would put up with that.
But yeah rotor run-out can wreak havoc on fixed caliper systems. It doesn't take much run-out to kick the pistons back, and then it takes extra pedal travel to affect braking. If you are aware of this extra pedal travel (never mind the hammering going on),and you don't do something about it: don't be blaming the system.

Part deux;
So lets talk about drum brakes;
Lets say the front W/Cs are 1-1/16, area is .8866 each. Lets guesstimate that the total piston travel on one side, is .150(yes I'm guessing) so the volume is .8866 x .15 =.133 cubic inch; add the otherside and you get .266 inch/cubic;or 4.36cc. This is 4.36/1.102 =about 3 times as much as the disc requirement.
Do you still wanna argue that a drum-drum M/C won't work on a disc/drum car?
I wish this idea/argument would die.

Do the math, find exact figures, prove me wrong. You won't hurt my feelings.I will say sorry, if I erred. There's a few guesses in here.