Furcurequs (aka Dwayne) questions Jim_Mich

[eccentrically1]

here is a paper that analyzed a 1/10 scale model glider in a wind tunnel at Langley circa 1955:

http://ntrs.nasa.gov/archive/nasa/casi. ... 026291.pdf

there are a lot of radius of gyration formulas. those guys seem to understand what they mean, or there would be a lot more accidents, or, they're just lucky.

just my opinion.

[Grimer]

Dwayne
...
Regardless of what you think, I believe Jim is an asset to this forum, even though I am unconvinced of his CF/momentum energy creation theories.

+1

He's certainly drawn my attention to the importance of Ersatz Gravity. :-)

[nicbordeaux]

Without Jim, this forum would be dead. He is entitled to his views about PM, energy creation, and else. He could be right, he could be wrong. So what ? OK, Jim does sometimes formulate things in a slightly peevish manner, but heck, any posts I don't like, I just ignore.

[Furcurequs]

You know, it seems I'm doing a lot of the work here.

jim_mich, your little cut-and-paste responses for some reason reminded me of this. Perhaps it is a bit more your speed, anyway. Enjoy. Oh,and what's a little ribbing between friends, huh?

http://www.youtube.com/watch?v=Cs4Gj7JsET4

By the way, the offer I made still stands.

If I'm such a thorn in your persecuted side, you can be rid of me with but a single post, you know, right?

Just plug some numbers into those formulas of yours to show that your method for solving that problem indeed works as you have said it does - and you can be rid of me forever. (Well, unless you later see me in the public eye, of course.)

Are you going to do that?

Have you tried to do that already?

Can you not actually do that?

...hmmm...

Did you not tell us the truth? Did you perhaps speak an untruth? You wouldn't have teased me or anyone else along with any sort of claim that you hadn't first verified, would you have?

...hmmm... ...again...

Could I possibly be hearing you sucking it up now?

Everyone else, I apologize for not getting back to you. Health problems and all. I needed a bit of a breather, too. I'll try to get back to posting in this thread sometime later. ...unless, of course, I have to check jim_mich's math first. Oh, and if that happens and the answers are correct, I guess I will at least post to say "bye".

If jim_mich doesn't or just isn't able to work his math problem, however, so as to get rid of me, I guess we'll then be able to hash out all kinds of important issues, won't we?

Take care, everyone.

Dwayne

[rasselasss]

http://www.youtube.com/watch?v=MAfCQ-t7xY0

[Furcurequs]

rasselasss,

Since there may be a tiny hole in your bucket (and, yes, I realize this will lead to my frustration after seeing your video), let me explain this to you in simple terms.

jim_mich accused ME of being a liar and so I had to show him that I was telling the truth and then he turned right around and - as the evidence would seem to indicate - LIED TO ME!

I'm thus willing to sacrifice my very presence in this forum just to find out the truth.

Did jim_mich turn right around and lie to me after accusing my true words of being lies?!

If he can show me that his words are true like I showed him that mine were, then I'm out of here. ...oh, and I'll even add in an apology.

...but if he cannot or will not do that and then has to continue playing games or making excuses, then he is not an honorable man and he is a major hypocrite.

It will not be a liar or a deluded man that will get you to the promised land. Even I can promise you that.

Take care (and maybe even be careful who you choose to follow).

Dwayne

[rasselasss]

http://leadershipfreak.wordpress.com/20 ... -im-right/

[jim_mich]

From: http://www.besslerwheel.com/forum/viewt ... 402#104402

Weights on rod = 1kg each
Rod weights are 1m from the center.
Pulley has a radius of 10cm
Falling mass = 500g

Is there a formula that will tell me how much time and how far the weight would need to fall to reach 50RPM?

Radius of Gyration formula:
RoG = (a²×W+b²×w)÷(a×W+b×w)

W = 2000 g (two 1_kg weights at ends of rod)
a = 100 cm (radial distance of weight 'W')
w = 500 g (dropping weight)
b = 10 cm (radius of pulley)

Filling in the values:
RoG = (100²×2000+10²×500)÷(100×2000+10×500)

Radius of Gyration result:
RoG = 97.804878 cm

In other words, the arrangement will accelerate and rotate as if it were a single 2500g weight located at 97.804878 cm radius, sans gravity.

The 500 kg dropping weight is at 10 cm while the RoG is at 97.8 cm, so the torque force is leveraged.
It will be 97.804878÷10 = 9.7804878_g of force acting on 2500 g of mass.

The desired speed is 50 RPM. The RoG is 97.804878.
Note that Radius of Gyration value must be used as the radius of the moving weight, else the calculations WILL NOT BE CORRECT.
I've seen some members just grab any old radius that they come up with from who know where.
If you want accurate calculations concerning rotating objects then you MUST reduced everything that is rotating down to equivalant Radius of Gyration values.

So the circumference at RoG is 2×pi×97.804878 = 614.526172727 cm.
The desired velocity is 614.526172727 × 50 RPM = 30726.3086364 cm/min.
Or 30726.3086364 ÷ 60_seconds = 512.10514394 cm/s velocity desired.
Or 5.1210514394 m/s velocity desired.

We want to know the time needed to accelerate 2500 grams up to a Radius of Gyration velocity of 5.1210514394 m/s using a force of 9.7804878 grams where the gravitational acceleration is 9.80665_m/s².

Time = (M × V) ÷ (Ga × F)
Time = (2500_g × 5.121_m/s) ÷ (9.80665_m/s² × 9.7804878_g)
Time = 133.48 seconds to accelerate the mechanism up to a speed of 50 RPM.

Final velocity of dropping weight = 50_RPM × 10_cm × 2 × pi = 3141_cm/min = 52.36_cm/s
Distance dropped = (V×T)÷2
Distance dropped = 52.36_cm/s × 133.48_s ÷ 2 = 3494.5_cm or about 114.65 feet.


Hopefully I've made no mistakes. I make no guarantee. I've checked and double checked and believe this is the correct answer.

I made one small error...
The line that reads:
It will be 97.804878÷10 = 9.7804878_g of force acting on 2500 g of mass.
Should be:
It will be 500_g × 97.804878÷10 = 51.1222_g of force[/color] acting on 2500 g of mass.
By mistake I failed to multiply by the 500_g.

And of course this error trickles down through the rest of the calculations.
The bottom line results should be a time of 25.5367 seconds to accelerate the mechanism up to a speed of 50 RPM.

As a double check I just now went back and simulated it using WM2D, which came up with an answer of 21.358 seconds. The discrepancy is that with my calculations the weight masses out on the ends of the rod are fixed to the rods and thus are rotated, whereas the weights of WM2D were allowed to free-wheel and thus not rotate. Fixing the weights to the ends of the rods rather than letting them free-wheel results in it taking a slightly longer time to accelerate up to speed. In this case about 4.17 seconds longer.

I suppose I could redo the WM2D simulation and fix the weights so they don't free-wheel, but I've already wasted enough time on this.

This method, which uses radius of gyration works, contrary to what Furcurequs claims.

From now on I'm never going to try to help anyone, for if I make a mistake (we are all human) then the trolls here crucify me.

[jim_mich]

jim_mich accused ME of being a liar and so I had to show him that I was telling the truth and then he turned right around and - as the evidence would seem to indicate - LIED TO ME!

I don't think I ever accused Furcurequs of being a liar.
A did a forum search of all my postings for the following terms and came up empty:
Furcurequs lie
Furcurequs liar
Furcurequs lying
Dwayne lie
Dwayne liar
Dwayne lying

So Dwayne, if I ever accused you of lying, please post a link to such.

[Fletcher]

Jim_mich - your above WM pic is labelled Fc1 result [file shows as Ec1.wm2d in pic] whilst your file download is called Fc1.wm2d - whatever, FYI, the WM file doesn't download except as a page of gibberish.

[jim_mich]

I'm not sure why Firefox shows it as gibberish, except that you are looking at the raw info of the file. Do a SaveAs download instead, which seems to work OK. This is the first time I've uploaded a WM2D file in many years.

And yes I changed the file name after doing the screen capture. I realized after first making the files that naming them Ec1 rather than Fc1 might cause confusion with eccentrically1.

[Furcurequs]

jim_mich accused ME of being a liar and so I had to show him that I was telling the truth and then he turned right around and - as the evidence would seem to indicate - LIED TO ME!

I don't think I ever accused Furcurequs of being a liar.
A did a forum search of all my postings for the following terms and came up empty:
Furcurequs lie
Furcurequs liar
Furcurequs lying
Dwayne lie
Dwayne liar
Dwayne lying

So Dwayne, if I ever accused you of lying, please post a link to such.

You spoke of my words as being lies. That's calling me a liar, right?

So, maybe it was worded "that's a lie" or something like that. I'll look and get back to you when I can and I will also look over your math and check it against the other methods.

Dwayne

[Furcurequs]

It seems that I'm going to be having company tonight, so I will probably not be able to get to this until later in the weekend. Sorry about that.

Dwayne

[pequaide]

I have rechecked my data and calculations and they are correct. You can not do your caculations by taking the applied force out to the accelerated mass, you have to do it the other way around. You have to calculate what the accelerated mass would act like at the pulley where the applied force is. This would give you 20 kg at the pulley; accelerated by 500 g. 500/20,500 * 9.81 for an a of .292 m/sec² at the pulley. The final (target) velocity would be .512 m/sec at the pulley. And 5.12 m/sec (Your target) at the accelerated mass.

The two kilograms has a leverage advantage of 10 so it will act like 20 kilograms at the pulley radius.

The .500 kilograms exerts 4.905 newtons of force. So you have 4.905 = 20.5 kg * a; or .239 m/sec/sec.

The final velocity is 50 RPM / 60 sec = .833 RPS * 3.1415 * 2r = .523 m/sec

d = ½ v² / a; so d =.5 * .523* .523 / .239 = .573 m

d = ½ at², so it will take = 2.19 seconds

I have a wheel that I am now playing with where the wheel itself probably has a mass of about 1.8 kilograms. I have 3.8 kilograms on the circumference as in an Atwood's. The bearings are used and at times you can feel them grinding. The rolling resistance is probably 40 grams (force) or so. The drive mass is at .5 radius and the drive mass is 150 grams. Subtracting the rolling resistance the actual drive mass is about 110 grams.

This wheel will make one rotation in 5.5 seconds. This is a final velocity of .319 m/sec for the drive mass. And the drive mass will drop about .88 meters.

The drive mass of the real experiment is about the same (.110 kg at .5 and .500 kg at .1) as your thought experiment but the driven mass is at least double and you have a real bearing for the 5.5 seconds.

It seems that the calculation of 2.19 seconds for your experiment is consistent with real data.

Calculations are for estimating the results of an experiments. Taking the 2 kg into the pulley works; moving the force out to the mass does not work. The 150 g dropped .88 m is a real experiment and it confirms the 2.19 second range not the 21.3 sec.

[preoccupied]

Jim_Mich wrote:
From now on I'm never going to try to help anyone, for if I make a mistake (we are all human) then the trolls here crucify me.

You are welcome to help me whenever you like. Your opinions are also welcome.

I had a wild idea in my thread about motion without external force that had no basis except that swinging my arms around made me feel like something was happening. I'd enjoy knowing what you think about that. I started with my fists together and swung out my arms to a right angle and back again and it felt as if I was pulled forward, or there was at least a much sharper jolt to my mid section than the other combinations of swinging my arms around. So I propose based on my arm swinging experiment sensations that oscillating two weights on levers should be tested to see if they can move a system without external forces. It's not the most scientifically based proposition but I really seriously felt like I had much more impact to my mid section to the right angle starting with my fists together than any other of the combinations I tried. It reminds me of the string of beads that can fly over a jar when held high enough. I could be fooling myself.

amazing bead chain experiment slow motion
http://www.youtube.com/watch?v=6ukMId5fIi0