How to estimate the impact force of a martial artist breaking bricks? I was watching a martial artist breaking a stack of 9 out of 10 concrete bricks. I noticed that the last didn't break. It made me wonder if this would allow calculating the force dealt to be simplified.
The line of thought being it was enough kinetic energy for 9 but not 10, giving a limit. Rather than if it just went all the way through, you'd only know the minimum dealt force, but not maximum. Upon traveling down this rabbit hole I realized just how many variables there are.
So my question is, if I wanted to calculate the force dealt, what would the formula look like?
I understand there are different ways to attack this, but there exists some things I'm not going to know and that could be used to hopefully refine the angle of attack. For example I can weigh a brick, but I have no idea how I would measure how much mass he put behind the blow, so we can rule out the F=M*A approach.
edit: First, thank you for the responses. I appreciate any and all input to work towards better understanding on this. This is my first time on this site and based on the answers it seems my expectation of there being more of a conversation working towards an answer was incorrect. I also think I was expecting/hoping for something along the lines of "oh you just need to know the Joules needed for one brick and multiply it out because of x,y,z.." With all that said Ill gladly supply as much info as needed for an answer, or an idea of an answer. 
-the bricks were spaced about half an inch by small wooden pegs at either end
-L:15" W:7.5" H:1.5" Weight: 50lbs ~mass: ~22.68kg 
-I don't know how this would help but it seemed the strike starting from rest traveled ~4.5ft in ~.5s

Two blocks each of thickness h acting independently
[note: Ze is Z sub e, Zc is Z sub c, My is M sub y, Ze is Z sub e, b is width]
Elastic Modulus of a single block(beam)= Ze=I/y=(bh^2)/6
Combined elastic Modulus of two block acting separately= Zc=2Ze=(bh^2)/3
Bending Stress f=My/I=M/Zc=(3M)/(bh^2)
f=My/I=M/Zc=(3M)/(bh^2)
Now if the there was a single block of thickness 2h,then
Elastic Modulus of a block having thickness  2h: Ze=I/y=(b(2h)^2)/6=(2bh^2)/3
Bending Stress f=My/I=M/Zc=(3M)/(2bh^2)
Thus double bending stress will be generated in separated blocks as compared to non separated, thus Half force will be required for separated configuration in comparison to not.
This can be easily extrapolated to n number of blocks.
Edit again: really the biggest question here is: Does the fact that the last brick doesn't break reveal some key to the force or energy dealt? BUT a formula for the scenario with explanation of how you derived would be very enlightening and.. well fun
 A: That's a really interesting question!
I don't think the formula would be linear. I think it will take a lot of energy to break the first brick and then some of that energy would continue to be transferred through the later bricks. i.e if it takes 20N to break the first brick I don't think it will take 100N to break five bricks stacked on top of each other. (the blog post below disagrees with this)
I found this pretty cool blog article if you want to check it out 
http://www.dontow.com/2008/06/the-physics-of-martial-arts-breaking-boards/
He says that it depends on how they're stacked.
"Bricks, on the other hand, are ceramic, and snap (or shatter) upon a large enough impact force. Bricks don’t flex. When a stack of bricks is unpegged, the amount of force required to break all of the bricks increases with each additional brick. When bricks are pegged, the gap actually allows the bricks (which are heavier than the pine boards) to break each other, i.e., the force shatters the first brick, and the brick pieces falling downward through the gap will break the second brick, and so on. The larger the gap, the easier the following bricks will break, because the broken brick pieces will be falling for a longer time, picking up higher velocity and therefore leading to a larger impact on the next lower layer of brick. This means that breaking a pegged stack of bricks is significantly easier than breaking an unpegged stack of bricks."
edit: I see some of the comments are saying the question is unclear and are a bit hostile, I do think it's unclear but I think Nam knows that there are a lot of variables and is interested in the physics. Maybe we should be a bit friendlier to a new member?
A: Having done a bit of brick- and rock breaking in my youth, I know a trick that makes it much easier: Support the brick or long rock by one end, so it floats about a quarter inch or more above a hard surface.  Give it a sharp blow in its middle, and the floating end hits the hard surface and the rock snaps in its middle.  Just support it by its two ends and it's much harder to break.  So, in a stack of bricks the challenge is likely to be just to break the top brick.  When that gives way, the ones below it should break pretty easily due to the sharp impact from the brick immediately above.  
If you want to address the question from a purely theoretical perspective, I think there's not enough information given to provide an answer in the form of an equation.  Probably you would learn a lot by trying a few simple experiments:


*

*Stack one brick of the sort used in the demonstration on two  half-inch wooden dowels, place a 2" diameter wooden dowel (a rough substitute for the heel of a martial artist's hand), and apply weights to press down on the 2" dowel until the brick breaks.

*Stack the broken brick on another brick, like the bottom two bricks in the stack, separated by 1/2" dowels.  Apply weight over the break in the top brick until the second brick breaks. 

*Stack two unbroken bricks, separated by 1/2" dowels and add weights pressing down on a 2" dowel in the middle of the top brick; see if the bottom brick breaks.  Repeat using smaller diameter separator dowels.  Try different separator dowel diameters and see if there is a minimum separation at which the bottom brick breaks.


Where the results of those experiments would lead, I don't know.  But I suspect that the separation distance is crucial to the number of bricks that can be broken with a given blow.
