KaiMartin wrote:
- ... each side needs to take about 20 Newtons of drag. This is equivalent to about 2 kg of mass hanging at half the length of the spar.
Bob Kuczewski wrote:I'm not familiar with that relationship. Is it a rule of thumb or is there some more explanation that can help show that relationship?
These sentences do a two steps:
- Convert a force to an equivalent mass given. Consider an object with a mass of 1 kg statically hanging from a thread. The object will pull the thread due to gravity. The precise amount of force depends the circumstances. E.g, it would be a lot less on the moon and 1kg of helium would even pull upward under standard atmospheric conditions. However, if we assume a location on the surface of mother earth and a fairly high density, the 1 kg object will pull on the string with 9.81 Newtons. I only wanted to give a rough estimate. So I rounded the relation to 10 Newtons per kg.
- For the sake of the argument I'd like to compare drag to a weight hanging from the spar at a single point. Actual drag does not apply to a single point but is distributed with varying strength all along the length of the wing. However, I implicitly simplified the situation by assuming the same amount of drag everywhere. Now I can combine drag portions of the outer wing with a portion of drag from a corresponding position on the inner wing. The torque of all of these pairs with respect to the root of the wing is equivalent to the same force applied to a point at half the spars length.
---<)kaimartin(>---