drone with variable pitch propeller

[marcofa]

After an hour and a half of headaches... I bought a saw and broken a shovel unnecessarily? It would seem so!
I also started like this, but experts in another forum had convinced me.
I won't sleep tonight. right punishment!

 

[marcof]

I also started like this, but experts in another forum had convinced me.

See what happens to dating bad companies? :tongue:

 

[marcof]

It's obvious that the center of gravity moves in two cases.

eh thanks to the piffer, so you have moved the baricentrto /quote]


giurin giurella that I had not read your post when I sent my:redface:

 

[Fulvio Romano]

at lightning.
the reason for which integration even if rudimentary is necessary in the fact that the force of each section goes with the square of the speed, then 10 gr to 100 mm will exercise a force enormously less than 10 gr places to 400 mm from the axis.
Try to count in two cases and you're gonna be scared.

without bothering to make calculations with discreet masses, which among other things is likely to solve only a particular case, we make an analytical calculation.

the centrifugal force (and don't break me! :smile: that we are on a non-inertial reference system) of the dm shovel element will be:

WOMAN

if we integrate it along the shovel will come:

(r dm)

because the angle speed is constant compared to the radius and can be brought out from the integral.

at this point, that integral, if divided by the total mass, is the formula of the centerpiece. therefore to use the center of gravity is no more or less close to reality, it is analytically correct.

I think those on the other forums got confused with the moment of inertia. there, the turning radius is different from the center of gravity, but because the square size (the distance) is a variable to be integrated and is not constant.

Right?

 

[essebi]

according to me though dm = area(r) * dr to be integrated for r=r_iniz, r=r_final, where area(r) is the section that depends on the radius... therefore w(r) does not come out of integral

 

[cacciatorino]

according to me though dm = area(r) * dr to be integrated for r=r_iniz, r=r_final, where area(r) is the section that depends on the radius... therefore w(r) does not come out of integral

no, while dm = rho * area (r) * dr

and in any case the integral of area(r) * dr is semlicemente the volume.

the correct reasoning would be:

dfc = dm * w^2 * r = rho * area(r) * dr * w^2 * r

== sync, corrected by elderman == [ area (r) * r * dr ]

 

[Fulvio Romano]

therefore w(r) does not come out of integral

w(r) sarebbe thing?

 

[cacciatorino]

w(r) sarebbe thing?

as if he were ancestors.... :biggrin:

 

[essebi]

Yes, well, I was wrong. I mean, wr^2 doesn't come out of complete.

 

[Fulvio Romano]

w is not function of anything, perhaps it is function of time, but we are integrating to the eulerian, so it is constant.

 

[cacciatorino]

Yes, well, I was wrong. I mean, wr^2 doesn't come out of complete.

If by w you mean angular speed, the correct formula seems to me to be w^2 *r, however r should remain inside the integral, as you said.