How Do You Perform Cross Linear Interpolation for Suction Line Pressure?

[mattskie]

This is not a homework problem, but...

I am given the equation of two lines at different temperature gradients for (suction line temp,suction line pressure) => (x,y)
At outdoor temp 95*F; y=.41667x+54.1667
At outdoor temp 85*F; y=.4231x+50.3846

I need to find suction line pressure for 72*F temp (72,y) at 92*F outdoor temp

I believe the method is cross linear interpolation, how do i do this??

Methodology needed, i can handle the calculations. This was the closest I've found to a method, but the notation is confusing and I can't get it to work: http://www.mathpages.com/home/kmath323/kmath323.htm

please and thank you

 

[haruspex]

I'm a little surprised that the link you provide defines cross-linear interpolation as taking the bisecting line in the x-direction and 'ordinary' linear interpolation as doing it in the y direction. I'm not saying it's wrong (I've never heard of the term and cannot find an independent reference), it just sounds odd. Either way, you're treating x and y unequally.
If you have no basis for treating x and y differently, you could do it thus:
- find the point where the two known lines intersect.
- take the line through that point which cuts the angle in the ratio of the temperature differences.
E.g. if the angle between the two lines is 10 degrees then take the line that's (95-92)/(95-85)*10 degrees from the 95F line.
In the present case there may well be an asymmetry since x is also a temperature.
Anyway, the method in the link looks straightforward to me. If it's giving you a problem, please post your working.