Incremental conversion to 24V possible?

I want to switch from 12V to 24V. I want to do this in an incremental way as much as possible. Components that need to be replaced (eventually) will be
 the PSU (most obviously)
 the heated bed and
 the heater cartridge
I plan on using a buck converter (that I have laying around anyway) to keep my fans.
Something upfront: I know that heated beds are not that expensive but this is mainly about doability of my idea/thoughts.
I now thought about the following. Would it be possible to only replace PSU and heater cartridge in the first step and keep the bed somehow? It is an Anet A8 12V 220x220mm PCB heater with aluminium heat spreader  that according to some sellers should be 24V compatible but I do not believe this.
I know, I could use the existing 12V PSU and a SSR to do this but I am seeking for a way that will not reuse the old PSU  if there is none that will also be a valid answer.
My first idea was that I knew from runningM303
you can limit the PWM factor to reduce the power. But then I remembered Ohm's law and this bed probably has a rather low resistance (I assume something around 1.5 Ohm), so this would lead to enormous currents flowing through the wires  even in case it is only on e.g. half of the time. So could I add another resistor to workaround this? Or any other way? 
You can only keep the bed heater ignition is dual voltage. A dual voltage bed heater normally has 3 terminations. For 12V operation , one wire is connected to 2 of those terminations and the second wire is connected to the remaining one. The connections can be changed for use with 24V.

@dc42 Thanks for confirming what I already expected.
I know about these dual voltage beds but I have no idea how they work. What is different when I connect one wire (usually positive) to two of the solder pads for 12V instead of only one for 24V?

@wilriker in the case of dual voltage heaters they normally have 2 resistance coils which are wired in Parallel for 12V and series for 24V

@wilriker
The wire on the heated bed pcb (or any other heater element) are behaving like resistors. If you are supplying power to a resistor, it heats up. Lets see some examples for a standard 12V heated bed:
(looks like this:)A    B
U = 12V
R = 1 Ohm (This is a standard resistance for a 12V bed).
I = U / R ==> 12 / 1 = 12 Amp of current flows threw the bed.
P = U * I ==> 12V * 12A = 144W heat is being generated by our bed.Let's see the same example with 24V supplied:
U = 24V
R = 1 Ohm (This is a standard resistance for a 12V bed).
I = U / R ==> 24 / 1 = 24 Amp of current flows threw the bed.
P = U * I ==> 24V * 24A = 576W heat is being generated by our bed, which is quadruple.
Now let's say, that the bed is actually 4 Ohm of resistance (this is a sandard 24V bed), but you put a third shoulder pad in the middle:
A  C  B
The resistance between A and C is 2 ohms, and B and C is 2 ohms aswell.
If you connect A and B to + and C to , then now you connected the two resistances (ac and cb) in parallel, so their resistance becames the half of each, so 1 ohms. 1 Ohm, 12V = 144W (see the calculation above)But of you connect A to + and B to  (and leave C unconnected) then you conected the resistances in serial, so their resistance adds up to 4 Ohm:
Let's see the same example with 24V supplied:
U = 24V
R = 4 Ohm
I = U / R ==> 24 / 4 = 6 Amp of current flows threw the bed.
P = U * I ==> 24V * 6A = 144W heat is being generated by our bed, which is the same as using the bed from 12V and connecting the resistances in parallel, So now your bed can be used in either 12 or 24V and generates the same 144W, you just have to rewire it to your needs. 
@denke Thank you for this very detailed explanation!

@wilriker You are welcome