Power Supply Build for Router CNC

[Jonathan]

The steppers will be wired parallel which the datasheet gives as 4.2A giving a combined draw of 16.8A. Using an approximate real draw of 66% because of the phase shift gives 11.2A which I'll round up to 12A to account for the drivers and any losses.

As I posted here, it's actually often quoted as 1/sqrt(2), so 70.7%.. but that doesn't change your conclusion.

Working backwards from my 70VDC target gives a secondary voltage of (70/1.4) 50VAC which I plan on getting by wiring a 25+25 toroidal transformer in parallel.

You meant series!

The transformer will need to be able to supply 70*12=840W so I was planning on using a 1000VA transformer

The transformer output will be 50VAC, so 50*12=600W. Nothing wrong with getting the bigger transformer if you want plenty of spare capacity for more axes, but you would be fine with a smaller one, e.g. 750VA.

The bridge needs to be a bit of a beast so I was going to go with this 400V 35A part. At full tilt it will need some serious cooling (assuming my calculation is correct)...

The bridge rectifier only needs to be rated for the average power, not peak. As I posted here the power rating of the motors you've chosen is found from their rated voltage and current, then remember to multiply by two as it's a two phase motor. That's P=2*4.2*2.73=23W. You've got 4 motors so call it 100W. Recalculate for 100W and you'll find if you stick with the 35A rectifier you wont need a heatsink.
Also, the rectifier and transformer are cheaper at Rapid Electronics.

After rectification it looks like I'll be getting 67.8V.

Remember it'll vary with the UK mains voltage tolerance, namely +10%-6%. That's still within the drivers rating though, but maybe a bit close so you could get the 24+24v transformer instead.

For smoothing I'll be using some cheap and nasty 100V 10000uF capacitors I've found on eBay. My calculation shows I need 28,288uF of capacitance to achieve 7V of ripple:

C = (20 * 0.01) /7.07 = 0.028288F

As the capacitors don't list a ripple current I was just going to stick 4 in parallel to give me 40,000uF of smoothing goodness.

As above, the current is actually a lot less than the value you've used in the formula, so 40mF will be plenty. as with most things, you can get the capacitors for even less on aliexpress. You might end up with fakes though, hence I'm not going to advise using smaller capacitors!

I originally designed the system with always connected bleed resistors but I changed my mind to have a relay system. I'm in two minds which is best but I've got the relay version drawn up at the moment.

The way you've connected the relay wont behave as intended, since (assuming the motors aren't connected) the voltage across its coil will stay high, as the capacitors are storing a charge, so the relay wont switch. The relay will only switch and connect the bleed resistor when the capacitors have already discharged a fair bit, which may take a long time. That kind of defeats the object. I'd go back to leaving the resistors connected permanently, as what if the relay fails? You don't need to save the few watts it can gain by using a relay.
Other than that, I think your circuit is fine.