ThinAirDesigns
Graduate Poster
- Joined
- Nov 10, 2008
- Messages
- 1,058
He probably won't post for a while so that he not obliged to answer the question.
ROFLAO. Genius argument - say something that has already happened, won't. Consistent at least.
JB
He probably won't post for a while so that he not obliged to answer the question.
Actually, if you're doing the analysis in the "ground's frame", it is force X distance over the ground.
What he is doing is mixing up energy required to do the work with energy balance. These are two separate issues.
In the first case I have 920 J/s KE loss of the air, plus 483 J/s of effective power spinning the perfect prop = 156 N X 9 m/s
In the second case I have an energy transfer of 920 J/s from air to cart system, with cart system losing 818 J/s in heat, friction and aerodynamic turbulence, and gaining 102 J/s in KE.
Let's take an extreme case. Suppose you have two magnets stuck together. Each exerts a force on the other. You pick them up and walk across the room with them. Even in the ground's frame, did either do any work on the other? They exerted forces on each other, and they moved relative to the ground, but it seems obvious to me that no work was done.
What do you think?
If you jump in the air, you continue moving with the surface, it doesn’t move under you, because you cannot escape from the inertia and the earth’s gravitational force on you.
On a treadmill, once you start to move backwards with the tread, you also have inertia and if you jump up the tread does not move under you either, so it seems the same as the surface of the earth, [...]
[...] but it isn’t! What if you don’t start the treadmill, but jump in the air and then switch the treadmill on while you are in the air? In that case, the tread will move under you. Now, instead of landing full on the tread, just very quickly and lightly bounce up and down. If you do it right, you can hold your position relative to the ground, while the tread moves under you. If you have a treadmill, try it!
I will bet you any amount of money you cannot bounce straight up and down on any level road that is on the surface of the earth, and get it to move under you! (without any external force, such as wind force)
So there must be something different!
That something is simply not allowing the inertia of the treadmill to have an effect on you, by never allowing yourself to be dragged back with it in the first place.
Why do you think the cart must be held in place while the tread runs under it, to get it started?
Two reasons, both equally important 1) to prevent the tread from building up any inertia in the cart 2) to get the wheels spinning and the propeller thrust going so the cart will no longer need to be held by hand. Now you have a cart in what is called a latched state, running on the treadmill with no inertia from the tread. It will slide and bounce around in a way that you will never see a cart on any road do.
The treadmill is NOTHING like a cart on a road!
I think you must be pulling our leg.
That's like saying that since the the Toyotas on the car transporter are chained down to its deck, no work was done as they traveled cross country from the dock to the dealer.
PS: and the post you refer to is spot on,
It doesn't seem to be defined well enough to give any kind of answer. But in the cart's case, it is a trivial matter that in the FOR where the ground is stationary, there is a force of 156 N (or 160) acting on the cart over a distance of 9 meters in one second. So that is 1404 (or 1440) Joules of work is done on the cart.Let's take an extreme case. Suppose you have two magnets stuck together. Each exerts a force on the other. You pick them up and walk across the room with them. Even in the ground's frame, did either do any work on the other? They exerted forces on each other, and they moved relative to the ground, but it seems obvious to me that no work was done.
What do you think?
I think I understand the second case. Not the exact numbers, but the general idea, yes.
There is work being done on the cart at that rate.I don't understand the first case. What is doing work on what, at the rate of (156 N)(9 m/s)?
The effective power is derived from the "Engine Power" formula on that page previously linked. It is the amount of power effectively converted to thrust AFTER aerodynamic losses (turbulence, rotational losses etc.) are taken out.What do you mean by "effective power" and where does it come from? And why are the two things you add together, appropriate to be added together?
Lots of work was done overcoming air resistence and rolling resistance and stuff like that. But I'm not talking about that.
I don't think any work was done by the force of the chains on the cars. Was some done? The chains were under tension, but they moved with the cars.
What did you mean by this?
(Here's the post in question.)
It seems you agreed with me then, but now you don't.
![]()
-Say that three times fast....![]()
By what? The air, I suppose. And what else (if anything)?There is work being done on the cart at that rate.
The effective power is derived from the "Engine Power" formula on that page previously linked. It is the amount of power effectively converted to thrust AFTER aerodynamic losses (turbulence, rotational losses etc.) are taken out.
It's the effective work done at the prop, and it gets added to the KE loss of the air to account for the total work done on the cart above.
By what? The air, I suppose. And what else (if anything)?
I'm still not getting it.
Ok, the "effective power" is the power that's available to drive the cart forward, i.e., to overcome air resistance and rolling resistance and/or to accelerate the cart. But what provides that power if not the KE that's continuously being lost from the air? I don't see why it makes sense to add them.
ROFLAO. Genius argument - say something that has already happened, won't. Consistent at least.
JB
No.
I don't understand.
Lots of work was done overcoming air resistence and rolling resistance and stuff like that. But I'm not talking about that.
I don't think any work was done by the force of the chains on the cars. Was some done? The chains were under tension, but they moved with the cars.
Spork said:But to someone standing on the side of the road, the cars on the carrier are being pulled through the air at 75 mph - by a chain. That chain is clearly doing work on the car.
Different frames - different energy accounting
"I thought you were at ...."
" I don't don't do much aero these days"
Ask SZ.
It's just funny that some emailed me because Spork gave them the opportunity.
However, if you put the money in a brown paper bag...
OK, Humber, £1000 in a brown paper bag, if you demonstrate your propless cart advancing against a treadmill.
Now, PUT UP, OR SHUT UP!
OK, Humber, £1000 in a brown paper bag, if you demonstrate your propless cart advancing against a treadmill.
Now, PUT UP, OR SHUT UP!