Continued from :http://forums.ubi.com/eve/forums/a/t...268#8631002268
Absolutely. It is no different than the thrust formula.I don't believe that the "mirror image" thing is more than an analogy, any attempt at a proof or complete explanation of vortex behavior.
Thrust = (Thrust Horsepower * 325)/ velocity in knots
At 0 kts we have zero thrust mathmatically.
However, this is not the case. Every fan company would be out of business if this was true.
Instead, to describe the physical behavior we have set the usable thrust at zero for zero velocity. In reality, that thrust force is met by ground friction and braking force.
It the same as the concept that Bill was trying to get out of Holtzauge.
http://forum.12oclockhigh.net/...=70663&postcount=111The question of Infinite Thrust at zero velocity was the question I posed to help you understand the difference between Net Thrust in the THp equations and Total Thrust (max) available as measured, at rest, for the system. Net Thrust will then decrease as velocity increases from zero up to max V.
So, answer the question - when does the equation for Thp relative to Thrust and Velocity become valid in your workd Holtzauge? Relative to total positive Force in the horizontal plane for level flight?
10kts? 30kts? 100kts? - and why doesn't it apply in deriving a Force vector on the system at zero to low speeds? What law of physics makes it valid at higher speeds? (The correct answer for industry standard practice is 105 kts) but remember it is for Net Thrust.
When do you arrive at a Force value to equate to Force Available - Force Reguired to overcome Drag and accelerate?
AndyJWest: You have no idea how happy I am to find a fellow pastafarian here! May you forever bask in the benevolent light of his noodliness!
However, Beware of the wayward pastor of creationist aerodynamics who lurks in this forum. Do not fall for his tricks. He has yet to see the light and to be touched by his noodliness.....
PS. Loved the picture. How can anyone doubt when his noodliness manifests himself so clearly?
The apparent division by zero just means mathematically it needs to be approached as a limit case.
First note that:
T =P/v
is a simplified case assuming a stable system where no acceleration takes place.
hence in our instance:
v (velocity) = 0
and dv/dt (aceleration) = 0
and d^2v/dt^2 (rate of change in acceleration)= 0
now, rearranging and solving for P:
P = Tv
If v=0, then the propulsive POWER of the propeller/engine as a system is ZERO.
Getting back to T =P/v.
Given P = 0 it can be easily shown that for
the limit case as v->0 then T=0
(in non mathematical terms - regardless of how small v gets P = 0 and hence P/v is zero and THRUST remains ZERO)
HOWEVER even from commonsense, if its a stable system where v is constant at zero their can be no EXCESS thrust and the total EXCESS thrust vector must be zero.
Actually this kind of thrust formula is not intended for zero speed situation, there is other kind of formulas for that. And the proper form of it is actually:Originally posted by WTE_Galway:
The apparent division by zero just means mathematically it needs to be approached as a limit case.
First note that:
T =P/v
is a simplified case assuming a stable system where no acceleration takes place.
hence in our instance:
v (velocity) = 0
and dv/dt (aceleration) = 0
and d^2v/dt^2 (rate of change in acceleration)= 0
now, rearranging and solving for P:
P = Tv
If v=0, then the propulsive POWER of the propeller/engine as a system is ZERO.
Getting back to T =P/v.
Given P = 0 it can be easily shown that for
the limit case as v->0 then T=0
(in non mathematical terms - regardless of how small v gets P = 0 and hence P/v is zero and THRUST remains ZERO)
HOWEVER even from commonsense, if its a stable system where v is constant at zero their can be no EXCESS thrust and the total EXCESS thrust vector must be zero.
T = P*n/V
Where and n is propeller efficiency which of course have to reach zero when the speed reaches zero while the power remains constant. If one wants to calculate out the propulsive power ie thrust horsepower (THp), then it would naturally be P(Hp)*n.
Note here that in the quoted TOCH thread Bill was originally arguing that thrust does not decrease when the speed increases later in thread he changed his mind... read it all if you are interested (some 15 pages).