Oh, you can. Very much so. I'm just not motivated enough to find out how :P

Basically, you'd have to set up radial and angular velocity (and subsequently the forces applied to P) just so that at every complete revolution (i.e. 2? in radian), you have an increase of 1m in radius.

Trivial, but not trivial enough to find a quick solution ;)

you cant do it that way. the particle isn't moving radially outwards at all moments. you have to set it up in n,t acceleration. that's normal tangential. i'm starting to look at my old notes now.

[edit] oh, it comes back to angular anyway it seems.

it can't be constant.
lets take some more conditions. If you want such a spiral:
r=1, theta = pi; r=2,theta=2pi, and so on. Thus, r = theta / pi.
and s = theta * r. so s = theta squared /pi
v = derivative of s. so v = two theta / pi
v = omega r, so omega is 2
normal acceleration = omega squared r.
4r is not constant.

i might be wrong, and these might be rules for circular motion only. someone plz check. i'm an electrical engineer, not mech.

If some undetermined force (let us call it n) that is variable is always perpendicular to x, (which it is, and it is) then the particle can do whatever the hell it wants. Since the slope of the particle at any given time is dependent on n, and since there's no way to nail down a definition on what n is in any which way, you can conclude that there's infinite many solutions.

Take a pencil starting at the origin, and draw whatever loopy curves you want, so long as each time you decided to cross the x-axis the line you draw is just one unit more along than the next, and you'll have your solution. And if you're not into n being a 'negative' force, then just always make c-clockwise spirals and you're golden.

Essentially you're trying to pin down a value on a variable which by necessity must remain inconstant, and can range from between -infinity to infinity at any given time (since you didn't specify how many times it needed to cross the x-axis.) Does that make me rich?

Edit: For example, this totally works. What n is equal to is just some very complex, very convluated value.

Let's attack this problem like a couple of engineers should: If r is the radius (in meters) and phi the angular coordinate (in radians), the conditions for the spiral would be:

r = r(phi(t)), phi = phi(t), i.e. r is a function of phi, and phi is a function of the time.

We know that F_r = 1 N (I am assuming smittens meant this to rotate, as well, not to let it point constantly in the direction of +x).
We also know that F_phi = const., but we do not know the base value.

We do know initial conditions r(0) = 0, phi(0) = 0.
We know that for phi = 0 + k*2?, r = k*1m.

If I missed any conditions, please point them out.
If you disagree with the ones I already stated, please correct them as you see fit, and I will respond to that.

Alright, majestic professional statuses aside, if the particle spiraled, then it probably wouldn't be crossing the x-axis 1m to the right each time, as spirals tend to visit each side of each axis in turn, including crossing such and such meters to the left on the x-axis.

You want to try something like I drew above, albeit more pretty, and with patterns like you physicist do. If you were cool, you'd have it hug the x-axis, and just cross with an infinitismal amount of each force at the right times. Such as...

t3h f0rc3 = ((x-0.00...01)(x-0.00...02)...(x-0.99...98)(x-0.99...99))((x-1.00...01)(x-1.00...02)... and so on.

or something so that the t3h f0rc3 in question is always zero, up until the point where x hits the 1m mark. I don't know if that's cheating or what in physics, as I never really took one of those classes, but whatever. I'm sure you can just just hack together a way of doing the same thing with your phi's, and i's, and radii or whatever you kids do these days.

We never said anywhere that it may not touch the other axes. I read smittens' post as a condition that each time the trajectory intersected with the x-axis, it should be 1m further along it.

Look bud, there's only one x-axis. If something spirals, it's going to touch the x-axis twice each spiral; once on the right, and once on the left.

If the spiral crosses *any distance* on the left side, ever, then it isn't crossing the x-axis "1m further to the right on the x axis" as per the condition. lolz! ¶}{¡zZ1x!

Point taken, he did indeed say that. He also said that it was supposed to spiral, though.

Maybe smittens could (finally) clarify what he really meant so we can get on with this.

Smittens doesn't know what the hell he's thinking; that's why he he's paying 2 mi to whoever does know what Smittens is thinking.

You know what he's thinking as much as I do.

The only reason you know what he's thinking is because I already told you the answer.

I think the only thing we've learned from this thread is that Smittens doesn't know how to describe or think through a problem properly prior to posing it as a question for others to solve.

Well, you're entitled to your opinion. Mind you, my explanation still hasn't been proven false.

Just because it lacks proof to the negative does not make it true, but that's just a reminder.

Never in this thread did I try to prove you wrong (unlike in others (: ), I was only interested in fiding alternative solutions that would function in the way I had understood smittens' post.

However, I think that for providing such a solution, not enough conditions are available, and we could only solve it if we were to assume simplifications.

Sorry for the delay in replying again to this thread...but you guys have totally lost me.

So here's what I want: A particle that spirals counterclockwise so that every time it passes the positive x axis, the point where it crosses is one meter further than the last time it crossed the positive x axis.

Okay, so now can you figure out how that all works?

And maybe this should be in community projects, since it is everyone working together to figure out what I mean :)

assuming.... that you mean you want it to cross one meter distance per cycle (and not every half cycle, once up, once down.), then the formula is:

z = revolutions per second
y = m/s (in the positive x axis)

z = y

Indeed. Except for the units, but I'll let it slide for today ;)
The forces required are equally simple to calculate, since you have increasing radius, and thus require an increasing centripetal force, for which the formula is

F_c = m*w²*r.

r = r(t), i.e. it changes over time. We do know that r(t = 0) = 0 m.
Let dr/dt = 1 m/s. Radial acceleration a_r has to be 0 (thus, there must not be a constant force applied in the radial direction. This is what threw me off, I think).

w is revolutions per second, which we want to remain constant. Let's assume 2pi/s (1 RPS), to keep it simple.

Thus, F_c = F_c(t) = m*w²*r(t). So, for any given time t, you can calculate the radius r(t), multiply it with m and w², which are 1 kg and 4pi*s^-2. With the unit of r being meters, this yields kg*m*s^-2, which is Newton.

Voilà, m'sieur. F_c is what you want. Calculation of the tangential force will be done at some other point in time, since I have to set about my work now.

Mind graphing it?

Use your imagination, Kerner.