These questions are rhetorical...
So in the above examples... have we not forgot that ORD probably has an elevation, meaning the 3000ft MSL will mean an altitude of 2800ft* ish to climb? (This may be food for thought only for "academics"/calculator-pilots).
*) Did not remember the elevation, but it was 200something.
Another thought... Is the DME adjusted to read ZERO at DER of RWY in use? If the DME reads other than ZERO at DER, how much will that affect the gradient?
And will SLANTrange versus geographical 5 nautical miles be an issue (give large errors in calculation) when we talk about a climb gradient of 10%?
I am pleased to read the rules of thumb regarding calculation of climb gradients. There is a lot of knowledge within this forum, no doubt!
Boeingisgoing:
And by coincidence it works out that a given climbgradient multiplied by 2 is the climb per minute in feet at 200 kts groundspeed.
In this example: 10% X 2 = 2,000 feet per minute at 200 kts groundspeed. And if you are doing 250 kts, add 25% which will be 2,500 feet per minute.
Well, it is an even greater coincidence that any climb gradient in per cent (%) multiplied by the ground speed (GS) in knots will give you the required rate of climb (ft/min). This will be an estimate within limits that will vanish in the effect of i.e. wind.
200x10% = 2000fpm, 180x3,3% (200ft/nm) = 594 (rounded 600fpm)
Back to the O'Hare example, it may be so that the sources for error I am referring to above may be so small that they are negligible, but I think this is the right place to ask such questions. -So being a relatively junior pilot, I dare ask these questions and wait for my head to be chopped off (which happens often at these forums).