Bjorn,
Here is what Rick claimed.
"Basic numbers for 10kts:
...
Total drag 1540N"
You seem to believe that the total drag was for 1 m/s ~ 2 kn (Rick clearly stated it was for 10kts) and calculated this. "At 5m/s (10kn), the required power is 5*1500 = 7500W or 10hp. Looks reasonable to me."
Yes 10hp is a lot closer than 2 hp I agree ^^
What you are really trying to calculate is joules and yes your math would be correct if your (Ricks) basic assumptions were correct.
Also kts (knots) are not equal to kilometers. Do your calculations for knots and you will be even closer to the right number.
The basic problem here seems to be the concept of work (hp). If I lift a kg up and then put it back down, I didn't do any work. If I am holding a kg weight, I am not doing any work. If I press against a wall with a lot of force (newtons) I am not doing any work.
Lift is force in Newtons, a dimensionless number. It has to be made a vector before it can be converted to work, which is what your equation above does Bjorn.
Talador
Who flies, but is always looking for a landing spot when something goes wrong. Mr. Chicken here. . .
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Posted by: taladorwood@yahoo.com.au
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