Subtracting 2 Dates And 2 Times From Seperate Boxes

Jus thought I'd offer this code to the forum as spent a long time finding a suitable solution..

SELECT DischargeTime.ID, [Arrival Times].ID, [Arrival Times].[A&EArrivalDate], [Arrival Times].[A&EArrivalTime], DischargeTime.[Date of outcome], DischargeTime.[Time of outcome], DateDiff("d",[A&EArrivalDate],[Date of outcome]) AS DateDifference, [Time of outcome]-[A&EArrivalTime] AS TimeDifference, IIf([DateDifference]>=1,([DateDifference]*(1/24)*576)+[TimeDifference]*(1/24)*576,([DateDifference]*(1/24)*576)+[TimeDifference]*24) AS hours3
FROM DischargeTime INNER JOIN [Arrival Times] ON DischargeTime.ID = [Arrival Times].ID;

Basically I had an admission date and in a seperate box an admission time and then I had a discharge date and discharge time again in seperate boxes. The information was spread over two tables hence the join. How I solved the problem was by gaining both the days difference between and then the difference in hours and then I created an iff function and played around with it till it worked. Dont really know how i got it working in the end but it does so hopefully this code will save some1 alot of time!

(I know that if id shown my dates as date/time in 1 box the problem would have been much easier solved, however I dont personally like displaying dates and times that way especially when you have inexperienced people entering data)

Find below a result from the query

DischargeTime.IDArrival Times.IDA&EArrivalDateA&EArrivalTimeDate of outcomeTime of outcomeDateDifferenceTimeDifferencehours3
1119/06/200715:0420/06/200712:00103:0420.93