Landgraf Karl von Hessen-Kassel

[rlortie]

Please correct me if wrong (never claimed to be worth a damn at math).

A twelve foot wheel (3.6576m) with a circumference of 37.67 feet (11.49069m) turning 26 RPM will have the velocity of 980.2 feet ( 298.765m) Per Minute.

That makes for a speed of 11.16136 miles per hour (17.96247 k/m)

Is this anywhere near Correct?

Ralph

[daanopperman]

Let's make the wheel 11.5 m dia .

11.5 X pi = 36.128 times 26 rpm = 939.328 m per min X 60 min = 56359.68 m/h = 56.35 km/h which is just shi of 40 miles per hour .

[agor95]

@rlortie

Same as you - with maths - just typed the numbers in google.

12 feet = circumference 37.6991 feet x 26 x 60 = 58,810.596 feet / hour

5280 feet in a mile

So 11.138 m/h

You are in the correct area.

With the 8 weights of 4 pounds traveling at that speed.
We have some real energy in the device.

I would lock it in a room if I were you.

Regards

[daanopperman]

Nothing wrong with your math

3.65 m wheel will have rim velocity of 17.88 kph , just depend on your calculator digit ability . 17.88 X 5/8 = 11.175 mph .

[ME]

You are both correct.

Corrections on speeds:
(Ralph) 17.96 km/hr (I get 17.925, close enough)
(Daan) The diameter was 3.6576 meter, but with your radius the difference should be 23.8 MPH

ETA: at 50 cm from the rim the Centripetal Force equals 1g.

[agor95]

@ME

Lucky I have axe heads on both ends of the rods :-)

[rlortie]

Let's make the wheel 11.5 m dia .

11.5 X pi = 36.128 times 26 rpm = 939.328 m per min X 60 min = 56359.68 m/h = 56.35 km/h which is just shi of 40 miles per hour .

You throw-ed me for a loop on this one! A wheel 11.5 M dia according to my conversion would make it 37.73 feet in diameter!

No problem! I am happy and thank all for the final consensus.

Ralph

[daanopperman]

If the time is taken to move a weight in a wheel revolving 26 rpm , then it boils down to a optimum size wheel , where the rotation and also the power is maximal , unless the weights is rotating in a separate housing on the wheel rim .

Whether the wheel energy is from within or from the outside , the energy density used in Bessler's wheel must have been enormous . If form the outside no pressure differential , nor temperature or cariolus could have turned his wheels , and let's not forget the guy next door .

If from the inside , only motion , centrifugal , gravity , vortices from a working medium , air , and of those the energy density is not adequate , static electricity , magnetism , the Bernoulli effect , sound , these are only a few of possible candidates , and then a neglected possibility would be the different expansion rates in a bimetal strip on which a weight was connected .

So if we have to choose , a true ppm , or not , the force from within have much more possibilities .

[WaltzCee]

I never considered the g-force. Wow.


http://www.geneinfinity.org/sp/sp_rotor.html

[daanopperman]

@ Marchelo
Yes someone made a mistake , a 10 m wheel will have a higher linear velocity than 28. mph
You must have skipped my post while you were replying .

@ Ralph ,

S.A. changed to the decimal system in 1961 , so I did not reply in fpm , but if my memory does not fail me , 1260 ft in one mile ? Imagine what Aldo Costa,'s wheel will do at 26 rpm .

[rlortie]

@ dann,

"1260 feet in one mile?": One mile contains 5,280 feet or 1.609344 (km)...

My next step is to calculate the circumference velocity on a wheel six feet in diameter (1.8288 m) @ 26 RPM.

I am interested in the latency factor in getting a Newtonian liquid such as water to traverse from 10:00 o'clock to 2 o'clock seeking its own level.

Ralph

[ME]

Daan, I make them too.
26 RPM at a distance of 5.75 m gives 35.02mph (versus 11.14 mph)
I'm biased to not like Imperial conversions, it adds unnecessary complexities.
So I have to look them up: 1 mile =5280 feet (at least: the English mile)

circumference velocity on a wheel six feet in diameter (1.8288 m) @ 26 RPM.

8.96 km/hr, 980.2 ft/mn, 5.569 mph, Centripetal@rim= 0.6912 g


H = Height in [Feet]
S = rotational Speed in [RPM]
V = Velocity in [MPH]
conversion factor 28.011= (1.609344*1000*100) / (Pi*60*2.54*12)

V=H*S/28.011

[rlortie]

I have posted this before, but will mention it again for those who may not be aware.

Having a problem converting imperial to metric?

I use a free conversion tool with a shortcut on my task bar. It is free and can be downloaded here:

https://joshmadison.com/convert-for-windows/

You can enter your numbers, copy and paste the conversion.

Convert runs quite well on Linux and UNIX using Wine. If you do this, you may need to change the tab layout to use a single row by going into Options > Preferences > Tabs, and clear the Multiple Lines option. ↩

Ralph

[daanopperman]

Ralph ,

Mabee I have mistaken yards for feet .

ME ,

Ewig dankbaar .

[agor95]

This thread is golden.

It's good to know how fractional parts of a number can really create big differences.

It's good to know we like to use one measurement British Imperial feet - bigger feet [only kidding S.I.]

Also there are clues in Bessler's drawing to keep the movements small around the centre see M135 and use mass on each end of rods see the external pendulums.