can please someone explain this circuit to me

Hello, I have a question about a paragraph what I have read in datasheet of OP400. the complete paragraph has quoted here plus the corresponding figure, but for those who want the complete datasheet here is the link:
https://www.analog.com/UploadedFiles/Data_Sheets/3803980939304OP400_c.pdf

Quote OP400 datasheet from Analog devices:
DIFFERENTIAL OUTPUT INSTRUMENTATION
AMPLIFIER
The output voltage swing of a single-ended instrumentation
amplifier is limited by the supplies, normally at ±15 V, to a
maximum of 24 V p-p. The differential output instrumentation
amplifier of Figure 6 can provide an output voltage swing of
48 V p-p when operated with ±15 V supplies. The extended
output swing is due to the opposite polarity of the outputs. Both
outputs will swing 24 V p-p but with opposite polarity, for a
total output voltage swing of 48 V p-p. The reference input can
be used to set a common-mode output voltage over the range
±10 V. PSRR of the amplifier is less than 1 uV/V with CMRR
(G = 1000) better than 115 dB. Offset voltage drift is typically
0.4 uV/C over the military temperature range.

Click to expand...

OP400 is a package of four OpAmps with just two supply voltage pins(±V) for all four OpAmps. I really don’t understand it, maybe because English is not my native language. How it is possible to have 48V pp voltage at output with just one ±15V supply. I am also don’t understand the circuit completely so I will be grateful if someone explain it to me so I understand how it works. I understand this much that first two amplifiers(u1, u2) just amplify differential signals because their non-invert pins are connected together by RG and for common mode they are just followers. But u1,u2,u3,u4 never cannot accept inputs beyond ±15 nor they can provide outputs beyond that, thank you in advance!

look i dont understand it but i will ask for it the faculity if u can wait 1 day it will b very good and i will answer if they understand it
sorry for not solving it now
bye

of course I can wait, thank you for your kindness.

Hi,
The OpAmps U1, U2 and U3 form an ordinary differential amplifier which is an amplifier with very high input impedance and very low output impedance; I think its familier, this part has output ±24 Vpp if the output is from U3 to ground.

U4 acts as an inverting amplifier with unity gain. ie. its output is shifted by 180° from the input or simply it inverts (multiplies by -1) the input. This way if the output of U3 is Vo, the output of U4 is -Vo relative to ground. Now, if we take the output between the terminals of U3 and U4 making U3 the positive terminal, the ouput will be Vo-(-Vo)=2Vo, this way the output will be in the range of ±48 Vpp.

In other words, the output is differential between two terminals, which are opposite in polarity, if one is 1V the other is -1V and hence the differential output voltage is 2 V. Hence if the maximum is 24Vpp for each OpAmp alone, the differential will have output between 48 Vpp

Added after 5 minutes:

Hi,
The OpAmps U1, U2 and U3 form an ordinary differential amplifier which is an amplifier with very high input impedance and very low output impedance; I think its familier, this part has output ±24 Vpp if the output is from U3 to ground.

U4 acts as an inverting amplifier with unity gain. ie. its output is shifted by 180° from the input or simply it inverts (multiplies by -1) the input. This way if the output of U3 is Vo, the output of U4 is -Vo relative to ground. Now, if we take the output between the terminals of U3 and U4 making U3 the positive terminal, the ouput will be Vo-(-Vo)=2Vo, this way the output will be in the range of ±48 Vpp.

In other words, the output is differential between two terminals, which are opposite in polarity, if one is 1V the other is -1V and hence the differential output voltage is 2 V. Hence if the maximum is 24Vpp for each OpAmp alone, the differential will have output between 48 Vpp.