shaunksb
Well-Known Member
It has nothing to do with friction, all you need is one real life aerofoil characterization graph. No models and no approximations.
Ahhhhh.....at last. Please continue or point me back to some of your earlier posts
It has nothing to do with friction, all you need is one real life aerofoil characterization graph. No models and no approximations.
I'll try to refrain from posting again until the official NALSA record has been set....![]()
The equations of course are no help to those who don't have the background to understand them
ubergeekian: "I'm sorry that Halfway hasn't felt able to answer the three simple questions I asked. Nevertheless, here's a simple one for you: relative to a reference frame fixed to the cart (or boat) how much kinetic energy does it gain as its ground (or water) speed increases?"
What an arrogant unforgiving, and completely wrong person you are. How can you assume I was even reading the forum.
As for your question. a fixed reference point on a cart means the cart CANNOT have any energy due to its momentum. Since it can not have any momentum. That is what you are missing all along. Your arguments are dogmatic and wrong. It can not have any linear or rotational momentum as it is fixed to the reference. So no energy can be gained or lost by the cart.
ubergeekian then goes on to confirm that he has not altered from the energy relative to the ground theory by saying "Air resistance increasing with the square of the apparent headwind, to a large extent. " So confirming his belief in friction being the fundamental limit.
Back to real world, so Ubergeekian can insult me again in my absence. (I notice he never answered the better specified situations either, I did actually answer his as being impossible.
ubergeekian said:
- A propeller is turning in still air. The thrust produced is measured. With the rotational speed and pitch held steady, a tail wind (ie from the direction of thrust) starts blowing. Does the thrust produced increase, decrease or stay the same?
- A helicopter is hovering in still air. Thermal activity starts and the helicopter starts to rise with the air around it. Rotor pitch and speed stay the same. Does the shaft power required to turn the rotor increase, decrease or stay the same?
- In the above case does the work done against gravity increase, decrease or stay the same?
Is this a Troll?
So the drag off the little carty thing is more than the drag of the great big propeller? So much more its actually has enough power to turn it the wrong way when moving off. That's a lot of drag!
The cart is "aerodynamic" because in the back he has got his dads starter moter hooked up to all hi mates ipod batteries with his mothers washing machine belt driving the prop forwards.
THE EMPEROR HAS NO CLOTHES.
My guess is that a variable pitch prop would allow a faster start, by setting the blades fully fine ( flat ) to get the maximum 'spinnaker' effect to overcome the start friction then feed in coarser and coarser pitch to accelerate away. A simple centrifugal pitch control would do it.
Ubergeek and SL, an aplogy.
Good, though lofty explanation here:
http://projects.m-qp-m.us/donkeypuss/wp-content/uploads/2009/05/drela_efficiency.pdf
faster-than.html[/url]
Its a nice piece of mathematics but as is often the case , if the basis is wrong the whole analysis is wrong. Early in the alaysis there is the line which states that the shaft power of the air propellor is:
Pp = Fp (V −W)/np
Where V is the wind velocity and W is the vehicle velocity. If they are the same then the apparent wind is zero and the shaft power is zero.. so where do we get the drive energy from?
Actually he's right. The vehicle starts to move simply by being blown downwind.
Two days ago, I didn't believe it- now I am convinced.
Its a nice piece of mathematics but as is often the case , if the basis is wrong the whole analysis is wrong. Early in the alaysis there is the line which states that the shaft power of the air propellor is:
Pp = Fp (V −W)/np
Where V is the wind velocity and W is the vehicle velocity. If they are the same then the apparent wind is zero and the shaft power is zero.. so where do we get the drive energy from?
On the treadmill the prop effects are on both sides of the prop. But in reality the down wind vehicle has not altered the air in front Answering a situation about stationary props in a flow does not come close to the situation when sailing or driving down wind faster than the wind. So restate it in a relevant context or ignore it.
Are we all agreed now that the treadmill demo is just a well known high school physics trick.
Are we all agreed now that the treadmill demo is just a well known high school physics trick.