I found Vout of just the wheatstone bridge part is:
Vin((10.01/20.01)-(10/20))
Using 5 as Vin i got Vout=0.00025V

Does the 0.00025 Volts go to the + terminal of both of the op amps?
I'm not sure how to solve from here.

I know the transfer function of the amplifier circuit:

is

would I use the 0.00025 as Vin2 and 0 as Vin1?

 

I would think that sounds right. I have been wrong quite often though

 

thanks, can anyone verify this?

 

forgot to multiply by 5, Vout for the wheatstone bridge is .00125V

 

The more I look at this, the more I think V2 should be 2.5, and V1 2.5012.

 

The more I look at this, the more I think V2 should be 2.5, and V1 2.5012.

ya but still when you out those into the final equation you will subtract them and end up multiplying by .00125

 

I would imagine in a real circuit implementation there would be a power supply common reference at some point - say at the lower point of the bridge. Otherwise one might have issues with lack of conducting paths for input bias currents - should they be non-zero.

 

Google -> instrumentation amplifier This will give you some tips. A instrumentation amplifier will amplify the difference between the two inputs.

 

It's going to amplify the difference, but as far as the formula goes wouldn't you use 2.5 and 2.5012 for the V in 1 and 2? I thought that was the original question.