On September 30 2018 13:18 FiWiFaKi wrote: If you assume the earth to be a perfect sphere, then the distance to the horizon would be the same in every direction, which is you can derive geometrically as:
d^2 = h^2 + 2*R*h
where: d = distance to horizon R = radius of earth h = height to your eyes
Now you'll take that distance, and calculate the area... You can treat the earth as flat at such small surface area calculations and induce almost no error, so:
Then just take horizontal field of view over the whole circle, so for humans roughly 210 degrees / 360 degrees.
A = pi*d^2*(human horizontal field of view)/360
Now that we're talking about a practically flat piece of land, you're looking more of less horizontally (a couple meters down over the distance of several km). Your eyes have a field of vision roughly 70 degrees down compared to what you're looking at, and since you're looking horizontal, technically you can't see a very small distance near you, tan(20)*h, which is negligible.... Therefore the area you can see is:
39.659 million m^2 or 39.659 square km or 426.89 million ft^2
(I used 1.7m for the height to a 6ft person eyes)
The distance straight forward is 4652 meters. The distance spanning the furthest point left to right will be twice that (9304km), as you're seeing a a fraction of a sphere, and since the 4652 meters is the radius, and you can see more than 180 degrees of the circle, the furthest distance between points would be the diameter 2*r = d.
So when something flies, it needs the same or a stronger force pulling the object down in order for it to take off (or stay in the air). Let's take a drone. When it's in a box (or a surface area) that's standing on a weight scale, how high does the drone need to hover before the weight being displayed on the scale starts to differ, aka be less because of kinetic energy being dissipated in the air and the container? Volume doesn't seem to be the most important factor, because the walls of the box can be down and hovering still seems to make it the same weight as the drone standing still. However, without walls, the height will start to have an impact much faster I think. Would anyone be able to give some kind of quantitative relation of sorts?
On October 08 2018 07:47 KwarK wrote: In a closed system a helicopter hovering inside a box would weigh the same as one landed because Newton. But that’s physics in a theoretical setting.
True. And for actual values as to how much of the lift pressure is put onto the box at variable heights in a real setting with an open box, you basically just have to test it. I don't think there is any reasonably simple theoretical way of solving this.
Your inference that a closed system is the same as an open sytem makes no sense. If the walls of a box was so closely contoured closed to the drone that no mass of air is being pulled down in the closed system, then there is no force acting on the drone fixed to the box but gravity. (Ignoring the forces of torque and possibly intake and outake thrust, but I guess you are concerning yourself with "up " and "down") The same applies if the rotorcraft was fixed in some manner to the box itself so it cannot take off.
If the box was open enough that air can move down the box and up the contours of the box with lesser vicousity effects then the scale will simply measure the Weight - Lift. (Ignoring close ground effect.) And it will continue to do so till Weight = Lift.
A hovering rotorcraft will have the same weight as a non hovering rotorcraft, but the lift forces are different. If a helicopter was sunk into a soft ground (a gigantic scale of sorts!) then as the lift force increases, so does the helicopter would proportionally rise from the ground till it is level with the ground level the ground was formerly before a helicopter landed on it at the very point the helicopter Weight equals its Lift. (Ignoring close ground forces again.)
I am pretty sure that the question is "Given a drone in an open box (or possibly on top of a flat surface) on top of a scale, how does the force which the scale displays change depending on the height at which the drone is flying above it"
If it is in an open box (Open where? The sides? The bottom? The top? It actually doesn't matter though.), that is the same as not having a box (ignoring close ground effects and viscous effects).
In which case the answer to your interpretation of his question would be that there isn't a difference because the drone is flying in the air with no interactions whatsoever with the scale. (Ignoring close ground effects and viscous effects) Which is so bone crushingly obvious I choose to believe that Uldridge isn't so bone crushingly dim.
Actually I decided to reread the question again word by word and got stumped by the first sentence.
On October 08 2018 04:17 Uldridge wrote: So when something flies, it needs the same or a stronger force pulling the object down in order for it to take off (or stay in the air).
Please allow me to reverse what I had just written about what I just chose to beleive.
On October 08 2018 04:17 Uldridge wrote:Let's take a drone. When it's in a box (or a surface area) that's standing on a weight scale, how high does the drone need to hover before the weight being displayed on the scale starts to differ, aka be less because of kinetic energy being dissipated in the air and the container?
Assuming that it isn't fixed to the box, the drone will still be in ground contact with the box before the weight being displayed on the scale starts to differ. As soon as lift force acting on the drone is not 0, the scale will detect it.
If the drone is fixed, the box and the drone can be regarded as the same object in terms of forces acting upon the box/drone.
BTW
On October 08 2018 04:17 Uldridge wrote: ...aka be less because of kinetic energy being dissipated in the air and the container? Volume doesn't seem to be the most important factor, because the walls of the box can be down and hovering still seems to make it the same weight as the drone standing still. However, without walls, the height will start to have an impact much faster I think.
None of these sentences make any sort of actual sense.
1) Kinetic energy wtf? You are concerned with mechanical forces, not energy. Energy shouldn't even come into it. 2) Energy is not created nor destroyed. If the drone hits the roof of the box, forces will act upon it and energy will be transfered, but energy does not simply dissapitate. 3) Volume of what? The box? Why wouldn't it be an important factor? Where does this assumption come from? 4) The drone (assuming it doesn't consume fuel) will always have the same weight acting upon it's body no matter whether it is off, on, or moving at maximum speed, as long as it is somewhere where gravity is constant.
Most of the post is just jumbled up nonsense and assumptions that are false. The answer will be simple, if only I can understand what he is asking.
______
Also:
On October 08 2018 07:47 KwarK wrote: In a closed system a helicopter hovering inside a box would weigh the same as one landed because Newton. But that’s physics in a theoretical setting.
In a closed system, where if the helicopter is free to move in the box the helicopter maybe have the same weight, and the box may have the same weight, and the air may have the same weight, but as the helicopter would be free floating in the box, only the weight of the box and air will be read on the scale and the weight of the helicopter hovering inside the box will be taken off the scale.
In a closed system, where if the drone is fixed to the box, the drone/helicopter cannot hover so the box and helicopter can be regarded as one body. The scale reading will not change (except maybe some vibrations due to turbulence).
Well, yeah, if you ignore all the effects that make the question interesting, the question is not interesting. That is obvious. This question is focusing on exactly those effects which you choose to ignore in your answer, and instead use your limited answer to insult the person asking the question.
This is similar to answering a question about the terminal velocity of a skydiver with "Well, if you ignore the air resistance, there is no terminal velocity, and you are stupid for asking such a stupid question."
If you have a drone hovering very slightly above the scale, almost all of the force will be displayed on the scale due to ground effects. If the drone is hovering 100m above the scale, basically none of the downforce will be displayed on the scale. Thus, it is quite reasonable to ask yourself how that distribution works, and if there is a simple way to describe this, as there are quite often simple ways to describe effects that people who are not very well versed in physics can't explain.
However, in this case, the actual physics are pretty hard, and none of the usual tricks to make hard physics problems easy work. Furthermore, there is not one simple answer, because the result is highly based on a lot of different specific factors, like the exact geometry of the box, the room you are in, the size of the scale, how the drone is built etc. And since the question is exactly about the strange middleground between the easy states, all of those can matter in unexpected ways.
What I am saying is that the sentences Uldridge has written is nonsense. He only asked one question that almost made any sort of sense and whatever he is asking, it can be solved by drawing a free body diagram in your head.
I didn't ignore all the effects that make the question interesting,, for at no point did he ask about ground forces, that was I myself adding it in for completeness. In fact his question doesn't seem to be about ground force at all, which is why I made sure to point it out and write to ignore it, same way I wrote to ignore viscous effects. But I included it so if he wanted to, he can research it himself because I think he is asking out of genuine curiousity.
But if you would include ground forces, the answer to what Uldridge has asked for does not change at all.
If the box is closed and the drone free moving then the scale would still detect a change as soon as lift forces occur i.e. as soon as the rotor spins even a miniscule amount (assuming absolute sensitivity).
If the box is open then the answer is the same as above.
If the box is closed and the drone is fixed to the box, then as one body, the scale will not change (except maybe some vibrations due to turbulence).
There's no change at all to the answers I have already given already by ignoring the effects of ground force. See, the question can be interesting, but Uldridge's post is so full of unbeleivable nonsense and assumptions that are all red herrings, that I still don't actually know what he is asking. Maybe he is asking 5 different questions, whereas you think he is asking just one about the effeects of ground forces, which was something I added for the sake of completeness. If you want to know the effects of ground forces, there's probably a million publications where you can find the relationship between lift force and ground force as it relates to height. Heck, if thats the question you want to ask simberto, I can probably give a reasonably thorough stab at it, but that does not appear to be what uldridge is asking for at all.
On October 08 2018 04:17 Uldridge wrote: So when something flies, it needs the same or a stronger force pulling the object down in order for it to take off (or stay in the air).
Please allow me to reverse what I had just written about what I just chose to beleive.
On October 08 2018 04:17 Uldridge wrote:Let's take a drone. When it's in a box (or a surface area) that's standing on a weight scale, how high does the drone need to hover before the weight being displayed on the scale starts to differ, aka be less because of kinetic energy being dissipated in the air and the container?
Assuming that it isn't fixed to the box, the drone will still be in ground contact with the box before the weight being displayed on the scale starts to differ. As soon as lift force acting on the drone is not 0, the scale will detect it.
If the drone is fixed, the box and the drone can be regarded as the same object in terms of forces acting upon the box/drone.
On October 08 2018 04:17 Uldridge wrote: ...aka be less because of kinetic energy being dissipated in the air and the container? Volume doesn't seem to be the most important factor, because the walls of the box can be down and hovering still seems to make it the same weight as the drone standing still. However, without walls, the height will start to have an impact much faster I think.
None of these sentences make any sort of actual sense.
1) Kinetic energy wtf? You are concerned with mechanical forces, not energy. Energy shouldn't even come into it. 2) Energy is not created nor destroyed. If the drone hits the roof of the box, forces will act upon it and energy will be transfered, but energy does not simply dissapitate. 3) Volume of what? The box? Why wouldn't it be an important factor? Where does this assumption come from? 4) The drone (assuming it doesn't consume fuel) will always have the same weight acting upon it's body no matter whether it is off, on, or moving at maximum speed, as long as it is somewhere where gravity is constant.
Most of the post is just jumbled up nonsense and assumptions that are false. The answer will be simple, if only I can understand what he is asking.
On October 08 2018 07:47 KwarK wrote: In a closed system a helicopter hovering inside a box would weigh the same as one landed because Newton. But that’s physics in a theoretical setting.
In a closed system, where if the helicopter is free to move in the box the helicopter maybe have the same weight, and the box may have the same weight, and the air may have the same weight, but as the helicopter would be free floating in the box, only the weight of the box and air will be read on the scale and the weight of the helicopter hovering inside the box will be taken off the scale.
In a closed system, where if the drone is fixed to the box, the drone/helicopter cannot hover so the box and helicopter can be regarded as one body. The scale reading will not change (except maybe some vibrations due to turbulence).
A hovering helicopter isn’t weightless. It’s just exerting enough force downwards to push itself up. The weight of the helicopter would still register on the scale, assuming a sealed box in which the air forced down registered.
-The open setting had to do with: if the drone is hovering close enough to the ground, it did the same thing as in a closed box setting, but for the closed system it didn't seem to matter how high that drone was. Are all interactions then completely elastic and if yes (or no), how does that exactly work? Does it make sense what I'm asking or do I need to rephrase that last sentence? -Mechanical forces they are then, my bad. I thought the energy used, which puts more force on the air molecules were an increase in kinetic energy of the air molecules (if that makes sense) -So then I was wondering if the height mattered more, even in a closed system (if you're hovering at 100m high in a 1×1×100 closed box, would the weight still be the same as hovering in a 1x1x1 closed box?). But I guess my nonsensical way of writing these down (check your physics privilege, DMCD), needs to be a bit more.. thought out. Thanks for pointing that out though, I'll pay more attention to it next time.
If you have a closed box, the force onto the scale is always the same (if you wait for long enough for the system to be stable. It might not be the same while you are moving upwards, and it might not be the same if you have just moved the top, but the air is still moving in a non-stable fashion). Because all of the force in that stable system is based in the gravitational force pulling drone + box + air downwards. If the drone is higher, there is more weird stuff happening with flow of the air in the box, but since none of that air can escape sideways (box is closed) and put a force onto another patch of ground someplace else, after a long chain of pushing air in different directions and trying to push the sides of the box outwards, it eventually pushes onto the floor of the box.
Regarding DMCDs critique of your terminology, it is a problem that to many people the terms force, energy, power and pressure (and sometimes acceleration, velocity, etc...) are pretty much equivalent, while they have very different, very exact meanings in physics. In a teaching environment, that is something that one has to constantly fight against, but in an environment such as this, one can usually figure out which of the terms above is the correct one to describe what they mean, but that puts additional weight onto the person trying to understand the question.
Is there any special way to dispose of cannabis plants? With the pending legalization here (8 days) I was wondering if there are rules/best practices for the disposal of the plant after the leaves are harvested?
Why are you harvesting the leaves and not the buds? Also, youd be wasting a fair bit by simple disposing of the plant remains instead of extracting from them the leftover active cannabinoids.
Good questions. I know nothing about it (clearly). I just heard that there are regulations is some districts about what to with the waste, and others their is not. And I was wondering if places that have legalized it have. Or if you can just compost it in your backyard or whatever.
A lot of the waste regulations relate to the leftovers from hydroponic setups iirc, I would think simple composting of plant material would be fine given the low concentration of the good stuff in the non-bud remains.