What is the mathematical relationship between Specific Excess Power and an aircraft's acceleration at a given point on an E-M diagram? If an aircraft is on the x-axis of an E-M diagram plot (zero turn rate) where the Specific Excess Energy (Ps) = 800, we understand this to mean that the aircraft has the potential to either instantaneously gain 800 ft/sec of altitude OR instantaneously accelerate.  If an acceleration rate is desired, this could be determined by converting the 800 ft gained (h) as potential energy to kinetic energy using an equation V = SQRT {2 x g x h}.  Substituting the variables, V = SQRT {2 x 32.2 x 800} yields 227 ft/sec gained each second (or acceleration of 227 ft/sec^2)  However, this is an unusually high amount of acceleration for even a high performance jet accelerating in straight, level flight (7 g's).  There is something clearly wrong with my logic/math but I can't figure it out.  Can someone offer the correct derivation?
You forgot that you accelerate not from standstill, but a considerable speed. So that equivalent potential energy change comes on top of the existing kinetic energy and makes the energy change look much smaller (resulting from the square of speed in kinetic energy). The correct calculation is
$$\frac{\delta E_{pot}}{\delta t} = 800 \;\text{ft/sec}\cdot g=(v_1^2-v_0^2)\cdot \frac{1}{2\cdot \delta t}$$
Your acceleration will decrease as speed increases. If I assume a speed of 800 ft/sec, the resulting acceleration is to a speed of
$$v_1 = \sqrt{v_0^2+2\cdot g\cdot h} = 831.55\;\text{ft/sec}$$
within one second which is 31.55 ft/sec² or just shy of 1 g. The initial acceleration is 1 g and decreases as the aircraft picks up speed. In mathematical terms:
$$\ddot x = SEP\cdot\frac{g}{v} = v_z\cdot\frac{g}{v}$$
When climb speed (or specific excess power) equals flight speed, both cancel, leaving gravitational acceleration equal to horizontal acceleration if all climb potential is used for increasing speed.