try this
Do you mind if I simplify this somewhat? Mr Trip suggests 60 tonne takeoff weight and for a still air ISA day at sea level V2 = 135 is probably the laptop answer. (That corresponds to about 11 tonnes fuel, 11 tonnes payload and 1600nm trip with "typical" reserves by the way). You should be able to climb say to FL370 in 115nm in just under 20 mins with adequate instaneous climbing reserves. Probably thus: 250kcas below 10,000, 290kcas above 10,000 until 29,500ft where speed becomes M0.76 all the way to 37,000ft. Say 2,900fpm to 1500ft, 3,900fpm thereafter to 10,000ft then falling to 1,300fpm at about 29500ft where 290kcas and M0.76 are identical. From there a steady reduction to 900fpm until nicely settled at 37,000ft and accelerated to M0.78 cruise. There are a lot of assumptions there and not a lot of good engineering grammar but I hope as a guide it isn't way too out for what you want. If you accept this then the formula for time increments "T" is:
T = [(h2-h1)]/[(RoC2 - RoC1)] * natural logarithm[(RoC2 - Roc1)]
If you are going to go into much deeper detail you probably need someone with a BLT (no, that is a Boeing Laptop Tool and not a sandwich) who is prepared to do a bit of tapping. Otherwise the maths can get very tiresome though not unduly awkward with a spreadsheet.
Any help?