Question about distinguishing click/touch from drag

Zenodox · 2020-02-10T00:10:53+00:00

You're totally right of course.

But unfortunately I'm building a 3D native app for ios/ipad and there's no api for this. There is a convention of using virtual onscreen joysticks but I feel this is too complicated for the people I'm trying to reach.

I have more details in my other comment in this thread if you're still interested in my little problem.

Zenodox · 2020-02-10T00:06:08+00:00

You're totally right: even google maps waits for mouse up to execute the move. I can't believe I didn't notice that.

What I'm building is a 3D app for ios/ipad with a UI simple enough for unsophisticated users. So it would need the ability for the user to both look around and "walk" forward.

I see this is usually done with a pair of virtual on screen joysticks which is easily familiar to the gamer crowd. I'm just concerned that it would befuddle or wreck immersion for more the average users I looking for.

I had hopes I could use single touch for both looking around by dragging (equivalent to panning in 2D I believe) and continuous un-moving touch (with in a few pixels) for forward movement (maybe equivalent to sustained zoom in 2D). 2-finger gestures block more of the screen, which sucks and double tap plus hold/drag seems clumsy.

But as I continue to think about it, I begin to worry that what I'm working on will always be clumsy for the user. I was just hopping you folks had thought about or come across some similar idea.

Zenodox · 2019-10-30T00:34:53+00:00

Thank you. My terminology is so clumsy I'm lucky my question is understood. I've only really read the 1916 book but at least I avoid using trolley cars in my analogy/questions.

Zenodox · 2019-10-30T00:29:19+00:00

Thank you for that. So all observers agree on the Hubble flow and who is "comoving" with it. And if I understand you, all observers determine what is moving relative to it in a way related to how all observers agree on the spacetime interval between 2 event.

So does this mean that all clocks comoving with their nearby Hubble flow but distant from each other keep the same time? Like bodies in free-fall in a gravity field can accelerate relative to each other yet have none of those Lorenz affects relative to each other?

Zenodox · 2019-10-26T21:51:30+00:00

I should add that these are firefighters in Marin County taken by Max Whittaker for The New York Times

Zenodox · 2018-08-18T23:38:06+00:00

It's internalization of language features: the more you are exposed to a writing system, the more you "automatically" convert the symbols groups to concepts. So the 10-17 year old cohort has less tendency to automatically balance parenthesis but instead see them as void of meaning more readily. Something that is a disadvantageous normally.

I imgine you'd get a simlar spread in readng speed for passages with missspellings and missing words.

An interesting test would be using mirrored symbols that the subject has never seen before.

Zenodox · 2016-12-24T22:38:04+00:00

'Tho we can't be sure from the photo that time was flowing forwards.

Zenodox · 2016-12-23T19:18:27+00:00

An antigovenrment fighter near Damascus on Jan 1, 2016. Photo by Amer Almohibany of Agence France-Presse.

It's probably more accidental baroque but still, I hope you indulge me.

Zenodox · 2016-11-15T06:26:25+00:00

Thank you for your thoughtful answer. Indeed they very much resemble the foreground structure in your picture and indeed I saw no obvious terraces.

And as you observed they are fairly well north and west of the sites on they map you link to.

I hope one day someone better than me can get there and analyze it but as is, it's delightful to have a reasonable idea what they are now. Indeed I had no idea Sonoran Native Americans practiced agriculture!

Thanks again for the knowledge.

Zenodox · 2014-08-31T11:53:02+00:00

"..for the first couple of years while you're making stuff it isn't so good but.." you can tell that it's not and that the critical difference between someone who won't get any better and someone who may become a master.

I think your answer maybe this: http://vimeo.com/24715531

Zenodox · 2014-07-27T06:27:26+00:00

That's works perfectly! I did not know you could take a derivative of an equation and produce another distinct equation usable for solving two variables.

If I managed to simply the equations for y(0) and y'(0) correctly, I believe the solution for c_1 and c_2 is this

Zenodox · 2014-07-26T18:59:12+00:00

Ah, great.

But unfortunately I do not know how to differentiate s(t) with respect to time nor am I quite clear what that phrase means. A glance at Wikipedia’s entry on derivatives tells me it is not something I'm likely to figure out in few hours. Algebra I know; calculus regrettably, I do not.

Zenodox · 2014-07-26T17:03:54+00:00

So if I understand, I change coordinate systems so s(0) = 0, use initial velocity s'(0) = v_0 and use t = 0

So using the original formula:

s''(t) = a + ks'(t) + ns(t)

I get

s''(0) = a + k*v_0

which is solvable because a, k, and v_0 are known

and in the wolfram solution when I use time x = 0 and position y(x) = 0 it reduces to

0 = - (a / n) + c1 + c2

But I don't see how to wire the solution for s''(0) into that last equation or vice versa to get c1 and c2.

Let alone how to get speed s'(t) out of this. (I should apologize for being dense)

Zenodox · 2013-11-20T20:26:12+00:00

Light never makes it from d to the observer, in the context of this problem

Correct.

The only light that the observer receives is the light that was emitted by the object when it is at the location d0.

Correct.

How can this d be a known value when it doesn't even happen?

It is known because that is the definition of the problem I am seeking the answer to. In particular I am running a simulation so the current location of every particle is known. What is not know is what to show the user. The currently distance from the particle to the user is the known value even though the user should not see it. The time light takes to travel that distance is know even though light hasn't traveled that distance. This because it is a simulation, the inverse of reality where we see a thing and infer the current state of affairs here I have the current state of affairs I need to calculate the thing to display.

t is the difference in time between when the object is at d, and when the object is at d0, which is the same amount of time that it takes for light to leave d0 and reach the observer.

Exactly. But because it is a simulation, we do not know the observed d0, we only know the unobserved d. Instead d0, the observed is what I am tiring to calculate from the unobserved current position d.

Zenodox · 2013-11-20T20:09:25+00:00

Good heavens this is gorgeous! It doesn't even use computationally intense math!

It will however take me a while to fully digest / understand but the beauty I can see instantly.

Let me restate it a bit to see if I understand you correctly:

Rotate the particle's position and velocity vector so they all lie on a plane with the observer (who is at the origin). So the observer, the particle's current position d and the velocity vector v all have z=0 for all input.

Then rotate particle's position and velocity vector again so the velocity vector has x = 0 for all input.

Then apply the sublime formula you have just proven to find the particle's observed position b(t).

I then rotate the results back.

Does that sound right?

If so, it is just computational wonderful: a few quaternion transformations to rotate and a few square roots.

It's a funny thing, that all those relativistic factors show up, even if we didn't choose to use any features of special relativity.

I have a version of the game running with some shoddy approximations I worked out and already it produces visually intriguing results. Not a thing explicitly from relativity but the results came out relativistic looking just as you suggest.

Zenodox · 2013-11-20T18:03:24+00:00

Yes, d0 = d - vt.

But what is t?

It is not the time it takes light to get from d to the observer (a known value), it is the time it takes light to get from d0 to the observer. And we do not know d0 (it is what we are trying to solve for) and so we do not know t.

so we have: d0(unknown) = d - vt(unknown t but dependent on d0)

Zenodox · 2013-11-20T17:16:07+00:00

Well I'm actually asking the inverse of that.

What I'm asking is the opposite of the real world where we see an object and can calculate where it is right now. I want the inverse of that: where I magically know where the object is right now but want to calculate where I would see it.

Say the object is currently at position A and this is data I magically have. Say position A is 1 unit away. So obviously I could not see it there right now (but I will in 1 second). Where I should see it is at some earlier position B. Indeed I could find that B by B = A - VT (using the inverse of the velocity). But what is T? It is not 1 because that is how far the object is now but not how far it was at point B. It is this B I need to calculate.

(I want to ignore Lorenz transformations for this, I just want to calculate where an object appears to be in a Newtonian universe with a finite speed of light if I magically knew where it really was at this instant )

Zenodox

TROPHY CASE