General > General Technical Chat
Newton's third law problem.
electrodacus:
--- Quote from: Circlotron on November 28, 2022, 09:17:48 pm ---A superconducting motor or electromagnet would continue to supply a non-moving force without any ongoing power being consumed once initially energised. For that matter, so would a permanent magnet. I have seen that “supplying continuous power without movement” argument used by the free energy from magnets crowd.
As far as your arm needing energy to produce a non moving force, that’s true, but you could also get that force by for example driving a wedge into a narrow space. Once you have stopped pressing it in you are no longer putting energy into it but it continues to exert a force against an object. Are we to assume it continues to dissipate energy into this object indefinitely? A force, certainly, but there is no movement, so where is the energy?
--- End quote ---
Opposing magnets and springs can store energy. So there is a potential stored energy. So that wedge you mention takes advantage of materials elastic properties.
You do miss the point that none of your examples will work without brakes.
You have a vehicle on wheels with no brakes that needs to keeps his relative position on the ground using an electric motor.
The main question is the one currently discussed with Kleinstein. That is how much power is needed to advance at any low speed against the wind.
There is an equation for that and we disagree on which one is the correct one.
If the correct one is the one I claim it is then that alone is enough proof that no vehicle can drive upwind without energy storage.
electrodacus:
--- Quote from: Kleinstein on November 28, 2022, 09:19:01 pm ---Driving 5 km/h against a 30 km/h head wind is relatively easy, if you have suitable low gears (e.g. montain bike). The difficult point is more to keep the balance at slow speed, if gusty.
5 km/h is about walking speed and when walking 30 km/h head wind is not hard either.
Providing 460 W is in contrary only possible for a short time for a top athlet.
--- End quote ---
I know all this about walking speed and the 300 to 500W sustained by a human is possible for at least a few minutes that is why I selected the values for speed.
The difference is just huge 460W to 560W when frictional losses are added is very different from 65 to 150W you may be claiming.
As an example that calculator will say 85W for 5km/h with no wind so your claim is that wind makes almost no difference when that is not the case.
Also your theory is not consistent.
For example if you release the brakes and let the wind pushing you what will the power provided by the wind be with wind at 30km/h ?
Changing gears will not help with amount of power needed.
If you change in a lower gear yes the force you need to apply at the pedal is lower but the speed will be higher so exactly the same amount of power will be needed.
You will say that since bicycle is at 0km/h the wind power available to accelerate the bicycle will be zero.
While I say is almost 300W witch again correspond to what you will observe in reality.
Nominal Animal:
Here's some of the actual physics related to this phenomena, for anyone interested:
Let \$v_v\$ be the vehicle speed with respect to ground, and \$v_w\$ be the wind speed with respect to ground.
For a vehicle that is in continuous contact with ground (through e.g. wheels), the amount of power available from the wind is
$$P_\text{in} = \frac{\kappa}{2} \rho A_\text{in} v_w^3 \tag{1}\label{1}$$
where \$\kappa\$ is the extraction efficiency, up to about 0.593 in theory (Betz's coefficient), and up to about 0.475 in existing devices; \$\rho\$ is the density of air; and \$A_\text{in}\$ is the cross-sectional area of the extraction device.
In addition to the ground friction losses, the amount of power lost via drag is
$$P_\text{out} = \frac{C_D}{2} \rho A_\text{out} \lvert v_w - v_v \rvert^3 \tag{2}\label{2}$$
where \$C_D\$ is the drag coefficient depending on the shape of the vehicle (varies between approx. \$0.02\$ for airfoils to \$0.5\$ for golfballs and similar), and can be different for \$v_w \gt v_v\$ than for \$v_w \lt v_v\$ (i.e. front-to-back drag coefficient can be and often is different than back-to-front drag coefficient); and \$A_\text{out}\$ is the cross-sectional area of the vehicle.
The only situation where using \$P_\text{out}\$ to model the amount of energy available from airflow to a vehicle makes any sense at all, is when that vehicle is in flight and only uses airflow propulsion, and \$C_D\$ is used to denote the energy extraction efficiency \$\kappa\$, noting that there is absolutely no physical reason or mathematical relation tying \$\kappa\$ to \$D_C\$.
For example, existing airfoil-type wind turbines, ignoring the tower, have \$A_\text{in} = A_\text{out}\$, \$v_v = 0\$ (stationary), and \$\kappa \gt C_D\$. In current commercial wind turbines, \$C_D \lt 0.1\$, with \$\kappa \gt 0.4\$, i.e. \$\kappa \approx 4 C_D\$. If they had \$P_\text{out} = P_\text{in}\$, they could not produce any power at all. It is the difference between \$\kappa\$ and \$C_D\$ that allows a stationary object to extract power from wind.
The reason there is no \$v_v\$ term (typo fixed) in \$\eqref{1}\$ is that the fact that the contact to ground, even if rolling, allows the mechanism to balance static forces with forces against the ground. A similar effect with the keel is extensively used in sailing; the keel having very little drag in the direction of travel, but enormous drag in the perpendicular direction, allowing the keel to be used to balance the forces exerted by the wind to the vehicle. In recent decades, hydrofoils are being used more and more, lifting the vessel upwards, reducing drag (\$C_D\$) due to smaller cross-sectional area \$A_\text{out}\$, without increasing the total drag much at all.
In very rough terms, a sufficiently clever land vehicle can balance the forces from the wind with static forces through the contact with the ground. Such mechanisms will have some small losses, but at least in theory, they can be reduced without any known lower limit.
While we currently believe the upper maximum limit for \$\kappa\$ is indeed about 0.593 as described by Betz's limit (at essentially \$C_D \approx 0\$), there are no known physical limitations for the ratio of \$\kappa\$ to \$C_D\$: that is just an engineering problem.
Indeed, the Mythbusters experiment by dimpling a Ford Taurus with over a thousand golfball-like dimples, increasing the fuel efficiency from 26 to 29 MPG, shows that many widely used vehicles have an unreasonably large \$C_D\$. This affects what humans perceive as normal, and results in quite normal but efficient mechanisms obeying Newtonian physics to defy human intuition. Those interested in such quirks, may find the misconceptions related to Bumblebee flight informative.
Kleinstein:
Mechanical power is force times (inner product if using vectors) speed. So at a low speed there is little power needed. How much depends on the motor and gearing. With a suitable low gear the power can be very low and the motor internal friction may be enough. To provide some braking (intentionally avoid the word power here). Common languish english seems to use the word "power" also for a few different things than the physical power.
One can use this also to convert physical power back to the force. So the force is power divided by the speed.
I think it should get obvious that the idea of an essentially fixed power even at low speed leads to a contradiction, as this means the force about doubles when you half the speed.
So there must be some wrong point leading to deverging large forces: The formula or power = force times speed is universally accepted, sometimes even used as definition. So the wrong point is assuing the essentially (vehicle much slower than the wind) constant power from the kinetic energy of the wind as the power needed to move the vehicle.
Kleinstein:
--- Quote from: electrodacus on November 28, 2022, 09:57:46 pm ---
You will say that since bicycle is at 0km/h the wind power available to accelerate the bicycle will be zero.
While I say is almost 300W witch again correspond to what you will observe in reality.
--- End quote ---
The mechanical power is force times speed. So how on earth can you get 300 W with 0 speed as one factor.
A limited force and zero speed naturally gives zero power - where is the problem ?
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