I don't think the idea that they would have the same delta bt graph really makes sense. Let's go extreme and then Just make up numbers ...(I'm a total amateur though but have been reading to understand fluid bed, so this concept was important to me) let's say that the beans are at room temperature, you drop in to a room temp drum and turn up the heat. Obviously the gas has to heat the drum, the air, and the bean simultaneously so the beans will have a fairly low ror, most importantly here the drum and, depending on design, air temp, would heat up faster than the beans. Thus, the ror of the et would be higher than that of the bt.
If we preheat to 300 degrees, turn off the gas and just leave it off, then at some point the bean temperature and environmental temp will stabilize and be equal, at first the beans will heat up very quickly But as they get closer to the et the ror will slow down. Thus, let's call that initial ror 60f/min. As the bt reaches 150degrees the drum is still much hotter than the beans but is losing heat and the two are much closer in temp, so now the ror will be lower, let's call it 40f/min
say we now turn on the gas, at bt 150f, theoretical delta bt 40f/min. If heat is then applied it must first start with the drum, which is now losing heat to the beans and getting heat from the gas. In this way you can control the descent from 40f/min until drop.
Let's say instead we charge at 150f. Just by natural logic we can tell that at the moment of drop the beans have to have a lower delta than at 300f charge. If it was 60f/min at 300 then it must be lower at 150. Thus, the ror might now be 45f/min. Continuing to apply heat must first heat the drum, and then the beans, thus the bt ror will not equal 60f/min, it just won't drop as it would with the heat off. stopping my long winded speech here, it seems that in order to reach the benchmark goals the delta graph will have to look different. As to how they slurp, I don't have enough experience to speak to that.
Yes, i you per this on an iPhone