Table S1. Parameters used for generation of equations compared to Kohn (1996)
	Bird (Kohn 1996)	‘Herbivorous bird’	‘Carnivorous bird’	
Input (moles)				
Mass (kg)	0.10	200	200	
Metab pre-exp	2.93	2.93	2.93	
Metab exponent	0.75	0.75	0.75	
Energy (KJ)	151	44 819.96	44 819.96	
O2 respired (moles)	0.34	100.84	100.84	
Humidity (%)	0.75	0.78	0.78	
Temperature (°C)	15.00	5.00	5.00	
Air exchanged 	181.2	53 783.95	53  783.95	
H2O conc, sat’d (mmol l–1)	0.724	0.389	0.389	
Air H2O (moles)	0.10	16.21	16.21	
Rel. digestibility	0.90	0.70	0.90	
Energy extr. efficiency	0.90	0.90	1.00	
Food carbo content (%)	0.80	0.80	0.00	
Food fat content (%)	0.05	0.05	0.20	
Food protein content (%)	0.15	0.15	0.80	
MJ energy/kg food	15.26	11.87	21.62	
Food ingested (g)	9.90	3780	2070	
Moles O2/kg food	12.80	12.87	2.80	
Food O2 input (moles)	0.10	30.62	5.22	
Moles H2/kg food	32.07	29.45	20.80	
Food H2 (moles)	0.26	70.06	38.81	
Free-H2O in food (%)	0.65	0.65	0.65	
Free-H2O in food (moles)	1.02	389.63	213.93	
Urea/Uric acid (moles)	0.00	1.07	4.48	
WEI (mJ/KJ)	0.20	0.20	0.20	
Water turnover (moles)	1.68	498.00	498.0	
Drinking water (moles)	0.30	22.09	229.05	
Total moles O2 input	1.15	345.43	335.66	
Output (moles)				
Faecal output (kg, dry)	0.001	1.133	0.207	
Faecal H2O content (%)	—	0.70	0.70	
Faecal H2O (kg)	0.002	2.643	0.484	
Faecal H2O (moles)	0.50	149.40	149.40	
Urea or uric acid O2 (mo)	0.01	1.61	6.72	
Urine (moles)	0.00	0.00	0.00	
Nasal H2O (moles)	0.14	40.69	40.69	
Skin H2O (moles)	0.59	174.30	156.87	
Sweat/(Seat + pant)	0.00	0.00	0.00	
Oral H2O (moles)	0.45	133.61	151.04	
Sweat H2O (moles)	0.00	0.00	0.00	
CO2 (moles)	0.31	94.83	79.95	
RQ (CO2/O2)	0.91	0.94	0.79	
Total moles O2 output	1.15	345.43	335.66	
Input compositions				
Air O2	15.1	15.1	15.1	
Meteoric H2O	–3.25	–22.00	–22.00	
Air H2O vapour	–13.24	–33.19	–33.19	
Leaf H2O	3.25	–15.88	–15.88	
Cellulose	30.25	11.12	11.12	
Stem H2O	–3.25	–22.00	–22.00	
Leaf/(Leaf+stem)	0.50	0.75	0.75	
Food H2O	0.00	–17.41	–9.84	
Food O2	30.25	11.12	7.01	
D Atm vapour – SW	–9.99	–11.19	–11.19	
D Food H2O – SW	3.25	4.59	12.16	
D Food O2 – SW	33.50	33.12	22.90	
Output compositions				
Body temperature (K)	311.15	310.15	310.15	
D Oral H2O – BW	–8.19	–8.38	–8.38	
D Nasal H2O – BW	–17.05	–9.64	–9.64	
D Skin H2O – BW	–18.00	–18.00	–18.00	
D CO2 – BW	38.65	38.83	38.83	
D Urea – BW	0.00	0.00	0.00	
				
Air O2 input	5.13	1522.76	1522.76	
Moles air O2	0.34	100.84	100.84	
d18O relative to meteoric water (MW)				
Air vapour input	–0.50	–90.73	–90.73	
Food O2 input	3.46	1013.99	151.58	
Food H2O input	2.51	893.91	1300.86	
Drinking input	0.00	0.00	0.00	
Sum input Ds	5.47	18.17.17	1361.71	
Moles MW	0.81	244.59	234.82	
D18O relative to body water (BW)				
CO2 output	12.35	3682.06	3104.29	
H2O vapour output	–7.78	–2324.60	–22.40.75	
Urine + sweat	0.00	0.00	0.00	
Urea	0.00	0.00	0.00	
Sum output Ds	4.56	1357.46	863.54	
Moles BW	1.15	345.43	335.66	
BW = (A+B)×MW				
A = (d18Osum input Ds – d18Osum output Ds +d18OO2, air input) /BW moles	5.18	5.4	6.02	
A+18.5* 	23.68	24.24	24.52	
B = Moles MW/Moles BW	0.71	0.71	0.70	
Calculated d18Op = (A+18.5) + (B×d18Ow)	21.38	8.66	9.13	
*The value 18.5 is used rather than 17.5 for the fractionation between phosphate and body water because the model was generated using values that were measured using the BiPO4 method.  Our data were collected using the Ag3PO4 method.  There is a documented ~1‰ decrease in d18O for phosphate generated between BiPO4 versus Ag3PO4 (Chenery et al. 2010).  

In Table A1, column 1 is taken from Kohn (1996) (with the exception of A+18.5 – see Table A1 footnote) and was modelled using different meteoric water composition and temperatures.  The larger herbivorous and carnivorous bird columns that are used as proxies for dinosaurs are modelled with lower temperature and higher humidity to mimic the Alaskan climate in the late Cretaceous (bold).  The equation was solved for different relative humidity values, a regression line was generated and that regression line slope was considered the coefficient for the humidity factor.  That coefficient was added into the d18Op equation at 0% relative humidity.  The equation was then solved for d18Ow rather than d18Op as in Kohn (1996).  So the equation at 0% relative humidity for herbivorous bird (dinosaur): is

d18Op = 0.71d18Ow + 35.28;	(A1) 

by changing the input humidity for a range of relative humidity values (between 0.5 and 1), we generate a regression for the humidity coefficient of

d18Op = –14.25h +19.704.	(A2)

When this is included in the equation at 0% humidity we get:

d18Op = 0.71d18Ow -14.25h +35.28.	(A3)

This equation rearranged to solve for d18Ow is used to estimate herbivorous dinosaur drinking water (see equation 1, main text):

d18Ow = 1.41 d18Op +20.07h – 49.69.	(A4)

The same procedure is carried out for carnivorous birds (dinosaurs):

d18Op = 0.70 d18Ow + 23.72,	(A5)

d18Op = –5.11h + 13.092.		(A6)

Combining the humidity slope with the equation at 0% humidity:

d18Op = 0.70 d18Ow –5.11h +23.72.	(A7)

Rearranging the equation to solve for d18Ow produces equation (2) (carnivorous dinosaur drinking water):

d18Ow = 1.43 d18Op +7.3h – 33.89.	(A8)