MULU operations are not constant time but depend on popcount. That has an easy work around just also do the MULU of the bitwise inverse, both instructions sum to a constant cycle count.

The brick wall I hit is related to the Macintosh SE ram being shared between CPU and video system. Every few cycles the CPU stalls instruction fetching, which turns the both MULUs together take the same amount of time each time into there is slowdown depending on the popcount of the first instruction.

I don’t know if anyone has any solution for this, there is prefetching of one instruction so all should be well, but it seems like the cpu stalls the in progress instruction if the prefetch is stalled.

A 2048-bit RSA key is 256 bytes. A 256-bit symmetric key is 32 bytes and ECDH keys are roughly the same size. The state size of AES is 16 bytes and its S-box is 256 bytes. Efficient implementations use only a few KB of state. The vast majority of cryptographic algorithms just aren't that big, relatively speaking; although as you mention, they are very computationally intense. Hash states are in the dozens to low hundreds of bytes. For TLS implementation, the record buffers are probably going to use the most RAM; 16K each direction + additional overhead.