I have a magic system with very specific physical rules (hence why the question is ), but I need help working out a specific scenario. Also feel free to suggest better terminology where it can use improvement; explaining this system has always given me difficulties.

Ground rules

It works like this: a witch can make the normal force apply without physical contact. For example, imagine that you are an evildoer walking up towards a witch you want to murder. If she is inside a building, then she can take the wall behind her, and manifest its physical presence in front of your nose. You'll bump into what feels like solid rock to you, and probably break your face.

And a witch can go on the offense too. She needs to make a slight hop, and then as she is about to land, projects her own body right on top of your head. If you didn't anticipate it, then the sudden impact of the mass of a human body on your head can break your neck. There's more sophisticated uses of this force; a skilled witch in active combat will seem to be doing parkour in the air whilst people around her are kicked without ever coming into physical contact - at least not in a way you can perceive with your eyes.

Other aspects and limitations:

  • This kind of sorcery only carries over the normal force on the super-atomic scale, not photonic interaction, chemistry, etcetera; so the projections are tangible but invisible.
  • A projected surface has all the friction of its source object (at least the amount of friction resultant from physical ridges in the object, not that from any chemical bonding or whatnot).
  • It does not create a vacuum where you project the item; a projection is an infinitely thin shell.
  • You can make interactions with the projections as if they were the real objects; project a small cup and you can fill the projection with water, and see water hovering in the air, whilst the source cup feels heavier.
  • Force on the projection can move and/or destroy the source object as if it were interacting with the source object. This also works vice versa.
  • The direction of any forces involved is never changed by the magic.
  • The projection ceases to be once a witch stops actively maintaining it.

Let me know if this makes sense so far.


Now, the specific use case I want to figure out is flight. One of the things a witch can do is seemingly walk on air, by actually walking on the floor below, which she projects higher up, right below her feet. That way she can create staircases anywhere.

But she can also change the scale of the projections. Surgery is made much easier by projecting an upscaled version of the patient's body, covering the projection with some sand so she can see it, and then manipulating the enlarged body with her bare hands, allowing high precision.

And that allows for a much cooler form of flight. A witch can walk on the air itself; if she projects an enormous surface of air below her feet. It should be a large enough surface that she can step on the air, push herself off it, and take another step with her other foot before the first foot sinks too far.

And because this projection is perfectly smooth (it is an artificial gathering of air along a perfectly flat, slightly inclined plane), it is frictionless. Once reaching cruising altitude (basically just high enough not to bump into buildings), a witch can lie down and glide on the projected air, and traverse the sky at hawk speeds. She only occasionally needs to get up and gain more altitude; even more skilled witches can use the (projected) wind for that instead and basically take off without expending any energy.

How large is the stairway to heaven?

The stepping process is the specific use case I need to sort out. The question is: in terms of horizontal surface area, how much air is needed to walk upon, particularly to lift oneself up? Think of it like the witch wearing enormous sandals, perfectly stiff and weightless, of such an area that you can step on the air the way you can walk on snow with snowshoes.

I want to sort this figure out because it will allow me to put an upper limit on the size of both projections and source objects. A formula with as input an arbitrary weight would be most useful, as I am also considering the witches building "boats" to fly on the frictionless projected air, and then I would need to know the required surface area to support the boat too.

  • 2
    $\begingroup$ For what it's worth, the magic system makes sense to me. It's quite powerful and versatile. While not unique, it's still quite distinct to most common magic systems that are based on energy manipulation or doing some metaphysical "code" action (speak out an incantation, do magical gestures, enact a rite) achieves a specific result. This magic system seems very similar to Mage: The Ascension (the Correspondence sphere) and Mage: The Awakening (the Space arcana) both are tabletop RPGs and allow for similar manipulation. $\endgroup$
    – VLAZ
    Commented Jan 4, 2021 at 11:55
  • $\begingroup$ @VLAZ Thanks for the references; I'll see if either of those RPG's have rulebooks online. And I'm not concerned with unicity, just with having fun :-) $\endgroup$
    – KeizerHarm
    Commented Jan 4, 2021 at 12:01
  • 1
    $\begingroup$ You can buy the rulebooks from DriveThroughRPG. Although, there might be simplified versions released for free, I'm not sure. I'd also advise you to check the same site - it also has a lot of the free demo materials. You can also check the White Wolf (publisher of both games) for more information: Correspondence, Space. The Sphere/Arcana is a bit broader but in both cases still deals with changing the apparent location of something. $\endgroup$
    – VLAZ
    Commented Jan 4, 2021 at 12:07
  • $\begingroup$ Asked on Physics.SE as well. $\endgroup$
    – KeizerHarm
    Commented Jan 4, 2021 at 14:14

3 Answers 3


Fascinating question! Your magic scheme seems to make total sense to me.

So the witch is exerting a force of 700N (about 70 kg seems reasonable? I've never asked a witch her weight, seems terribly impolite) on the air, and we need to find out what area of 'snowshoe' would resist that force for long enough for her to step upwards on it - she's always going to be falling, but needs snowshoes big enough that the falling is very slow.

When we say 'very slow', if she stands still, she'll constantly be accelerating under gravity: there's no-way that anyone could float by standing on a (flat, weightless) platform, no matter how big it was. Now I can walk up stairs at about 0.4 m/s. If the witch put in the same effort, it'd feel like walking up stairs - or on one of those stepper machines at the gym.

Now using the drag equation (https://en.wikipedia.org/wiki/Drag_equation), with air density of 1.2 and imagining the snowshoes as a thin disk which would give them a drag coefficient (https://www.engineeringtoolbox.com/drag-coefficient-d_627.html) of about 1.1 gives:

$700 = 0.5 \times 1.2 \times 0.4^2 \times 1.1 \times Area$

$Area = \frac {700} {0.5 \times 1.2 \times 0.4^2 \times 1.1} = 6628.788 m^2$

which is a snowshoe of approximately 92m diameter ($2 \pi r^2$)

If your 70kg witch was prepared to expend the effort of someone sprinting up stairs at (2m/s), she could fly on the level with only 18m snowshoes.

If she was super-powerful at projections, she could fly level using stroll-up-a-wheelchair-ramp levels of energy (slope 1-in-12, so lets say 0.0833 m/s) on snowshoes that were 440m across.

Of course it would take more energy to accelerate up into the air in the first place, and remember that stairway is always going to be like climbing up sinking sand.

Hope that makes sense...

  • $\begingroup$ So, if I understand it correctly, the 6629 m2 figure would apply for a witch performing twice the effort of a regular stair ascent? If so then I have my answer! :D $\endgroup$
    – KeizerHarm
    Commented Jan 4, 2021 at 15:28
  • $\begingroup$ And you can use this editor for equations. Write the equation in there until it looks the way you want, then paste it in an answer between $ dollar signs. $\endgroup$
    – KeizerHarm
    Commented Jan 4, 2021 at 15:32
  • $\begingroup$ I reckon so. Walking up a flight of stairs in a house (2.4m) in six seconds, effectively creates a vertical velocity of 0.4 m/s. At normal (sea-level) atmospheric pressure, that amount of effort would let her stay level in the air using projected snowshoes of 6628m2. $\endgroup$
    – zactionman
    Commented Jan 4, 2021 at 15:36
  • $\begingroup$ And thank you for solving my puzzle! :D $\endgroup$
    – KeizerHarm
    Commented Jan 4, 2021 at 15:50
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    $\begingroup$ $Snowshoe Area = \frac{10 \times weight }{0.66 \times v^{2}}$. And for a witch relaxing on top of her projection, $Terminal Velocity = \sqrt{\frac{10 \times weight}{0.66 \times area}}$ $\endgroup$
    – zactionman
    Commented Jan 4, 2021 at 16:11

Your magic system sounds like your witch can somehow blow a lot of air in a given direction to exert a force.

I think we can model it with the equation of drag force

$F_D \approx P_d \cdot A $

To sustain the body of a witch on a walk you would need to equate their weight on the surface of their feet.

Assuming the area of a witch's feet is $2\cdot 0.25 \cdot 0.10 = 0.05 \ m^2$ and their weight is 700 N, you would need a dynamic pressure of the order of $700/0.05=14000 \ Pa$

Note that this is assuming that the witch is standing. Should the position change, i.e. laying on the belly, that pressure would be harmful. On the contrary, going from laying to standing would require an adjustment to avoid falling.

If we calculate the dynamic pressure as $P_d=$$1 \over 2$$\rho u^2 $ we can get an estimate of the amount of air needed per second, as the volume of air contained in the footprint surface times the air velocity.

  • $\begingroup$ I don't quite understand your first paragraph. Yes, witches could climb to space, but as they get higher they would have to "gather" more air for their projection to support their weight, because the air is thinner. They would need larger and larger "snowshoes", which is what the projection would be analogous to. My question is the size of their snowshoes for regular altitudes; if you need to rest your mass on let's say the area of a football field, then I know that as an upper size limit for a moderately skilled witch's projections. $\endgroup$
    – KeizerHarm
    Commented Jan 4, 2021 at 11:57
  • $\begingroup$ And yes, for high-speed flight an inclined projection is needed - though only slightly inclined, since it's frictionless (although there's still air resistance from the horizontal movement). The level of incline depends on the direction and speed of the wind; going downwind could possibly take you up an incline. But I'll correct that in the question, thanks. $\endgroup$
    – KeizerHarm
    Commented Jan 4, 2021 at 11:59
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    $\begingroup$ @KeizerHarm for fun, you can have a witch "fly" on a broom...by actually sliding down a railing. So, they have a broom but actually travel down a set of stairs with the broom handle getting all the friction. They can similarly fly by surfing or skateboarding. $\endgroup$
    – VLAZ
    Commented Jan 4, 2021 at 12:10
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    $\begingroup$ @KeizerHarm, I have edited my answer $\endgroup$
    – L.Dutch
    Commented Jan 4, 2021 at 12:28
  • $\begingroup$ Thank you, but just to get this straight, there would be no air blowing involved. When you walk on the floor, do you make the ground push upwards? Ground carries people because it itself exerts the normal force (not an actual force but it's the idea), and that is dependant on your mass and the size of your feet. If your feet were as thin as needles you would sink through many surfaces. The witches here just increase the effective area of their feet so that they can step on something as thin as air.... $\endgroup$
    – KeizerHarm
    Commented Jan 4, 2021 at 12:41

This is a pretty complex question. The plates you're suggesting, being infinitely thin, would be infinitely light, so exerting any force to them would send them off at light speed. In this case, you have just air to keep them in place as you step on them.

Amount of Air

How much air you need depends on how fast you want to go, as I'll get into.

Overall, I guess you need at least the same weight of air as the weight you're exerting with your step. Every cubic foot of air weighs about 0.0807 lbs. So, for a 100 lb witch, you'd need 1,239 cubic feet of air, and more for jumping. But air is a gas, so if you put any pressure on it it'll compress/move....


If the witch can make a box of air, that might work, so the air can't escape. Since the box is weightless, it wouldn't cause the air to fall, under the weight of the witch was added. Of course, no matter how much air you use, you'd cause it to accelerate downwards, until it reached an altitude where it's at neutral or positive buoyancy despite its weight.

Actually, with a big enough box, you could basically make the equivalent of a balloon. But the issue with that is air is not lighter than air, so any added weight and it sinks. If you could capture specific gasses out of the air, that'd be different.

With a large enough air balloon, still, you could get a moment to jump from one to the other, the balloon sinking as you jump off of it.


So, when you jump off it, it'll be moved at the same velocity as yourself, so half your energy would be moving the box. However, it does have a lot more drag, being a giant box, so you'd get more than half your speed/energy jumping off of these things.

A larger box would mean more energy goes into your running, but you can never get 100%, so it just depends how fast you want to run up this staircase, as to how much air you want.

This would be quite an exhausting climb, as climbing ordinary steps is already great exercise, but you could do it.

Hope this was helpful.

  • $\begingroup$ The plates themselves would be massless, but the witch resting on them has mass so that keeps them in place; at least for long enough to step off one. With a large enough air balloon, still, you could get a moment to jump from one to the other, the balloon sinking as you jump off of it. That's exactly my question :) Witches would still sink through the air, but much more slowly slower thanks to the increased surface area and thus resistance. I want to know how big must I make the balloon so that stepping from one to the other is a feasible method of gaining altitude? $\endgroup$
    – KeizerHarm
    Commented Jan 4, 2021 at 12:50
  • $\begingroup$ @KeizerHarm I think I described the answer to that? It honestly does depend on how much of your energy you want to maintain. With the one I mentioned, at roughly the same weight of your witch, you should be getting ~>50% speed, by my estimate. So, 1,239 cubic feet of air in a box seems a good starting point for novice witches. Of course, that was a 100 lb witch for simplicity. $\endgroup$
    – Johnny
    Commented Jan 4, 2021 at 13:03
  • $\begingroup$ Ok, so if I understand that correctly, using such boxes as staircases would work but carry you half as high for the expended energy? Basically feeling like a staircase twice as tall? :-) That would be a good starting point indeed. $\endgroup$
    – KeizerHarm
    Commented Jan 4, 2021 at 13:06
  • $\begingroup$ Although I'm not sure a box model would work. A box with of the specified volume would be roughly 3 metres cubed. I imagine that for the brief period a witch rests on it, the air in the top would be compressed whereas the bottom half would barely be affected, and thus not actually be carrying the weight of the witch; reducing the effectivity. Might as well model the top part as a flat plane, and then the question is about viscosity (how much would the air sink/compress in the fraction of the second you step on it). It is getting more complex :D $\endgroup$
    – KeizerHarm
    Commented Jan 4, 2021 at 13:10
  • $\begingroup$ But maybe that compression resistance - whatever the physics term for that is - would be the value to pursue, rather than how fast the air gets out from underneath the plane, because we're talking about a very large flat surface and not much air could escape from underneath it in the fraction-of-a-second timeframe we're talking. I must do more research... $\endgroup$
    – KeizerHarm
    Commented Jan 4, 2021 at 13:13

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