See photos: I threw some spare parts together to try to demonstrate how it all works...
A 24Vac from the transformer is an RMS reading of it's peak voltages... which is about +34Vpk and -34Vpk, for a peak-to-peak of 68Vpp. That's probably where you're getting confused about half the input. It's still a 24Vac voltage, just half is positive and half is negative. At no time will you get a potential difference of both half-cycles. (In this particular example, anyhow.)
Photo 1 - Ac input. - A 12Vac input (I used a variac to take 120Vac down to 12Vac) and this is the waveform on the scope. It is positive half the cycle, and negative the other half. RMS reading is 12Vac. Notice that the peak voltage is ~18Vpk. Peak to peak is ~35.2Vpp. The little number 1 with the arrow (on the left of the scope) is the zero voltage point. The zero reference (ground) is always in the middle of this AC wave, hence positive and negative. We never end up with a 24Vac potential from ground at any point in time.
Photo 2 - Bridge output, no cap. - I connected a bridge recitifier to the 12Vac, and applied a load which drew ~1A. Notice that now all the negative cycles have become positive. Peak voltage is still ~18Vpk. We aren't missing any part of the waveform, all we've done is take all the negative cycles and rectified them to be positive. That's all the bridge rectifier does. We have also effectively doubled the frequency from 60Hz to 120Hz.
Photo 3 - Bridge output, with cap - I connected a 680uF cap (what I had handy on the bench) and now you see the minimum voltage has started to come up. Instead of dropping to 0V each cycle, now we're up to ~12V or so minimum. This increases our voltmeter reading to ~15Vdc. Notice the 4.8V ripple... that's about 26% of the peak, which means I would need to use a larger cap to get a decent DC voltage w/ low ripple, (I'd probably try a 1000uF cap for this 1A load.) That can all be calculated though. The more current (power) the load draws, the lower this minimum voltage will sag (with a constant capacitance.) If we were to increase the capacitance to 1000uF, for example, we'd raise our output DC voltage, since the minimum voltage will rise. I'd expect it'd go from 15Vdc to somewhere between 16 and 17Vdc.
Photo 4 - Bridge output, with cap, no load - Now I've disconnected the load. Now we have a DC voltage with near zero ripple. All we've done here is charge the cap up to the peak voltage level... since there is no load to discharge the cap, it just remains at 18Vdc, and each cycle will top off what little it loses through leakage. We will measure 18Vdc with a voltmeter, even though the input is 12Vac.
If we were to use an infinitely large bulk cap, we would also get 18Vdc with almost zero ripple, no matter how much power the load was drawing.
