I have done a spreadsheet, in regard to my wheel and I have found that when a wheel is built with a circumference of 28 feet, with 26 spokes and weights that are 2000lbs each, you can get the 1340 horsepower that is required for a one megawatt generator.
This is at the first setting. If you set the machine at the second setting , you can get more then twice the horsepower.
(EXPONITIALLY SPEAKING).
There are many ways to get the one megawatt of horsepower, which is the rearrangement of height, weight, number of spokes and offcourse the settings.
The setting means: How much does the weight move on it's spoke.
The first setting, equals one foot of movement.
VANDUGEGS
One Magawatt of Power
Moderator: scott
re: One Magawatt of Power
Do not trust his spreadsheet calculation. Posted below is a complete email from Darrell, sent before the nondiscloser was signed.
From : DARRELL VANDUSEN <------------------>
Sent : Saturday, December 4, 2004 6:28 PM
To : --------------------------
Subject : FORMULA
HELLO,
THIS FORMULA IS FROM A PHYSICS BOOK.
( WHEEL TURNED BY CONTINUOUS FORCE AT 3:00) 4.5 LBS
LET M = Mass of wheel
R = radius
K = radius of gyration
M = mass that is always at 3:00
LET Z = angular velocity
K.E. = Wheel + weight at 3:00
(1/2Mk.sqZsq + 1/2mr.sqZsq = x)
(X/33,000 = h.p.)
IT IS A NEW COMPLEX MACHINE; MAYBE IT NEEDS A NEW FORMULA?
THIS IS THE BEST THAT I HAVE FOUND, AND IT DOES IT JUSTICE.
REGARDS DARRELL
I have removed empty lines just for brevity, and have dashed out the email addresses. As you can see, he uses the wrong formula for calculating the rotational kinetic energy of the wheel, then divides that by 330000 to get horsepower. The problem is, the kinetic energy of the wheel is not the same as the output energy, and when you divide energy by 33000, you get energy, but in units that are 33000 times bigger than the original ones. You do not get horsepower.
From : DARRELL VANDUSEN <------------------>
Sent : Saturday, December 4, 2004 6:28 PM
To : --------------------------
Subject : FORMULA
HELLO,
THIS FORMULA IS FROM A PHYSICS BOOK.
( WHEEL TURNED BY CONTINUOUS FORCE AT 3:00) 4.5 LBS
LET M = Mass of wheel
R = radius
K = radius of gyration
M = mass that is always at 3:00
LET Z = angular velocity
K.E. = Wheel + weight at 3:00
(1/2Mk.sqZsq + 1/2mr.sqZsq = x)
(X/33,000 = h.p.)
IT IS A NEW COMPLEX MACHINE; MAYBE IT NEEDS A NEW FORMULA?
THIS IS THE BEST THAT I HAVE FOUND, AND IT DOES IT JUSTICE.
REGARDS DARRELL
I have removed empty lines just for brevity, and have dashed out the email addresses. As you can see, he uses the wrong formula for calculating the rotational kinetic energy of the wheel, then divides that by 330000 to get horsepower. The problem is, the kinetic energy of the wheel is not the same as the output energy, and when you divide energy by 33000, you get energy, but in units that are 33000 times bigger than the original ones. You do not get horsepower.
Disclaimer: I reserve the right not to know what I'm talking about and not to mention this possibility in my posts. This disclaimer also applies to sentences I claim are quotes from anybody, including me.
re: One Magawatt of Power
Prove that this design can drive itself first and then maybe you could talk about horsepower! This is starting to get silly.
Re: re: One Magawatt of Power
... I have to agree...ovyyus wrote:Prove that this design can drive itself first and then maybe you could talk about horsepower! This is starting to get silly.
projection without any proven substance to it... is just empty projection...
"A man with a new idea is a crank until he succeeds."~ M. Twain.
- ken_behrendt
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re: One Magawatt of Power
Some inventors think that the reason their device is not working is because it is just too small. I call this the "Ferris Wheel Syndrome".
If the operating principle is valid, then it should work whether it is Ferris wheel sized or small enough to fit in the palm of one's hand.
ken
If the operating principle is valid, then it should work whether it is Ferris wheel sized or small enough to fit in the palm of one's hand.
ken
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:
Vaver = -2(√2)πdωcosφ
Vaver = -2(√2)πdωcosφ
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re: One Magawatt of Power
more spacifically, i have given darrels details, phone and other things and reported him for fraud.
when the others testify he is gone.
cya darrel.
when the others testify he is gone.
cya darrel.
Re: re: One Magawatt of Power
This is very true!ken_behrendt wrote:Some inventors think that the reason their device is not working is because it is just too small. I call this the "Ferris Wheel Syndrome".
If the operating principle is valid, then it should work whether it is Ferris wheel sized or small enough to fit in the palm of one's hand.
ken
Bigger will make a wheel proportionally more powerful only because it weighs more. If a wheel is twice the size it it will be 8 times the volume and could be expected to produce 8 times the power.
But if it doesn't work in a small version it won't work in a large version.
re: One Magawatt of Power
I have been keeping track of the number of views in the fraud section.
Among the seven posts, their has been about 842 views, in the last 3 weeks. If you are interested in getting envolved, say so.
If you have any questions, just ask.
VANDUGEGS@YAHOO.CA
Among the seven posts, their has been about 842 views, in the last 3 weeks. If you are interested in getting envolved, say so.
If you have any questions, just ask.
VANDUGEGS@YAHOO.CA
re: One Magawatt of Power
SCAM ALERT - Darrell Vandusen (aka VANDUGEGS) is still fishing and waiting for more nibble$