Attempting to generalize Fred Lunnon's construction of the apparently-largest volume regular 4-simplex inside a unit-side 4-cube with edge length=sqrt(2)... from D=4 dimensions to dimension D = arbitrary multiples of 4.... achieving edge length = sqrt(D/2)...: In 8 dimensions, choose 8 vertices each of shape (1,1,1,0,0,0,0,0) according to a constant-weight code with Hamming distance=4, for example the cyclic cyc(11010000), then add a 9th vertex at (t,t,t,t,t,t,t,t) where t = (sqrt(19) + 1)/6 = 0.8931498242 achieving edge length >= 2 = sqrt(4). In 12 dimensions, choose 12 vertices each of shape (1,1,1,1,1,0,0,0,0,0,0,0) according to a cyclic constant-weight code with Hamming distance=6, specifically cyc(110011000010), then add a 13th vertex at (t,t,t,t,t,t,t,t,t,t,t,t) where t = (sqrt(37) + 5)/12 = 0.9235635442 achieving edge length = sqrt(6). In 16 dimensions, choose 16 vertices each of weight=6 according to a cyclic constant-weight code with Hamming distance=8, cyc(1001000000010111) then add a 17th vertex at (t,t,t,...,t,t,t,t,t,t,t,t,t) where t = (sqrt(17) + 3)/18 = 0.8903882032 achieving edge length = sqrt(8) OR alternatively, choose 16 vertices each of weight=7 according to a cyclic constant-weight code with Hamming distance=8, cyc(1101011000001100 then add a 17th vertex at (t,t...,t,t,t,t,t,t,t,t,t,t) where t = (sqrt(65) + 7)/16 = 0.9413911092 also achieving edge length = sqrt(8). In 20 dimensions, choose 20 vertices each of weight=9 according to a constant-weight code with Hamming distance=10, for example the cyclic cyc(00010001001010011111) then add a 21st vertex at (t,t,t,t,t,...,t,t,t) where t = (sqrt(101) + 9)/20 = 0.9524937810 achieving edge length >= sqrt(10). In 24 dimensions, achieve edge length>= sqrt(12) similarly using weight=10 or 11 (both work). In 28 dimensions, achieve edge length>= sqrt(14) similarly using weight=12 or 13. In 32 dimensions, achieve edge length>= 4 = sqrt(16) similarly using weight=14 or 15. CONJECTURALLY this keeps working forever because a constant weight code of weight=(N-2)/2 and length=N always exists when N is divisible by 4 (here "weight"=number of 1-bits in the N-bit binary word), having Hamming distance >= N/2. Actually, this is equivalent to the conjecture that when N is divisible by 4, an NxN Hadamard matrix always exists having (N-2)/2 "plus ones" in each row (rest minus ones). That probably is a well-studied conjecture, but my mind is currently blanking on that, would need to consult literature. This whole construction seems to be rather near optimal in the sense that edge length sqrt(D/2) only falls short of the optimal sqrt((D+1)/2), which is what is achieved using Hadamard matrix in the case where D mod 4=3, by a little. Hence the D-volume achieved falls short of the optimum "Hadamard bound" by only a factor (1+1/D)^(D/2), which is upper bounded by the constant limit (1+1/D)^(D/2) = exp(1/2) = 1.648721271 at most.