Wrong.
Which is correct
No you are just trolling
gives you the wrong value mate
Not even a little.
The vertical component doesn’t become the “leftover 50%” because there ISN’T a leftover 50% , you can’t calculate it like that.
The total ISN’T a simple 1+1.
Edit:
Hold for a minute, i’ll show the math
20000/2 is 10000 which is the value of the thrust for 60º, 20000·sqrt(3)/2= 17300 which is the value for the lift, and sqrt(10000^2+17300^2)= 20000 which is the “groos trust vector”
he said the remaining 50% is giving it horizontal thrust
genuinely you are trolling at this point. IG you two are friends.
resultant length: 20,000
Creating two vectors where one is 60 degrees from the resultant.
Component a = cos(60 deg)*20,000=10,000
Component b = sin(60 deg)*20,000=17,320
a + b = 27,320 (which is NOT the length of the initial vector)
See how the horizontal and vertical vectors DOESN’T total to the length of the initial vector?
So the horizontal component being half of the initial vector DOES NOT mean that the vertical component is also half. Something he HASN’T said anywhere.
Lol what?
From this point (bold), in my opinion, the discussion went off in the wrong direction. The guy just threw away half of the engine’s power and kept the wing loading unchanged. So where did the vertical component of thrust go?
he said it was and he made the assumption that at 60 degrees of nozzle angle half of the power was going down and the other half was pushing it forward.
Literally just look at the source the power is split evenly forward and downwards if the nozzles are at 45 degrees.
Where did he say this?
The screenshots you have shown he does not say this.
split evenly is not the same as half of the initial. they are very different things when working with vectors.
It pushes the plane up
He does not say split equal, he says half of the initial. Two VERY different things.
This is literally what the chart shows. Why is it 10,000lbs at 60 degrees in the chart?
That’s exactly what I’m talking about, and it’s completely missing from FeetPics calculations.
yes it is
COS(45 degrees) = .707
so are you telling me a 45 degree nozzle position has a 70 % offset in 1 direction? (of the aircrafts vertical and horizontal planes)
.707*28,000= 19,796 lbs force 1 way
and 8,204 in the other
at 45 degrees of nozzle angle
sure…
A 45 degree offset produces a thrust vector 0.707 timers that of the total thrust.
the horizontal will be 70% of the total and the vertical will also be 70% of the total.
It’s quite literally the reason that bridges look like this:
1 Like