Precisely.
So if you have one vertical vector with length 2 ( v=(2,0) ) and one horizontal vector of length 2 ( u=(0,2) ) their resultant would not have the length 4, it would have the length of the vector w=(2,2) which in this case would just be sqrt(2^2 + 2^2) ≈ 2.83
And in this case its horizontal and vertical so they ALWAYS meet at 90 degrees. Meaning that simple sine/cosine math works just fine. Because the resultant will always be the hypothenuse of the two vectors right angle triangle.
Yes its already known that sine in particular is what is used to determine the final vector regardless.
If you recall what I said earlier simply dividing the nozzle angle by 90 to get the approximate resultant vector.
This holds up and will only result in a small percentile change when compared to the correct vector (60 degree nozzle angle approximation is only 3% different then the corrected math) and magnitude. As the resultant vector does not have any of the Pythagorean math added too it. Seeing as the resultant is actually the Hypotenuse of the 2 vectors if they meet at 90 degrees.
I did this so its just easy for anyone to get an approximate answer.
It was completely unnecessary to dive into any of this to find an average to begin with. (There was also a chart that told him exactly what the thrust split was for the selected nozzle angle)
So again slapping COS(60) into a calculator because the nozzle angle was 60 degrees isn’t correct. Seeing at the vector of the nozzle angle at 60 degrees it no longer acts horizontally and vertically.
The problem isn’t about vector addition, its that the guy just took the horizontal component of thrust and started calculating with it, ignoring the vertical component and its effect on the aircraft. Maybe I’m missing something too, but it looks far from the truth.
Yes, and if you do the math for every angle of that chart the thrust resultant is always 20,000 lb
Yes it does? How does it not?
very simply
seeing as he based the Trig on the 60 degree vector and not considering that the plane is not acting on a flight path 60 degree rotated nose upwards.
Your putting the angle in the wrong place.
Right, by the connection of the Hyp and Adj relationship in the COS
It was just to demonstrate how the COS(60) degrees doesn’t mean anything in this specific set up.
SIN needs to be used here anyways
But it does though. I’m really having trouble seeing how you are thinking here.
Doesn’t matter, Sin(30)=cos(60).
The only thing that doesnt match is the lift at 4º which is probably a typo
If this was true his calculation would have been correct
instead he got the resultant vector of 45 degrees Oj at 60 degrees Oj
That tells me the math didn’t math.
Where did he get this?
He simply slapped in 60 degrees of nozzle angle as if it was the COS(60 degrees)
with that 60 degrees of nozzle angle resulting in a 50/50 split
this is wrong
No no, the cos(60)=0.5 is only for the horizontal component, the vertical component would be sin(60)=sqrt(3)/2 ≈ 0.866
He hasn’t stated that the other component also would be 50% of the total.
Which is correct.
So why is is resultant vector so drastically incorrect then.
Because his two vectors are not acting in the same relative plane as the aircraft vertical and horizontal component.
You can literally picture this in your head mate.
Is it? What are you comparing it to?
are you being serious or just trolling, genuinely.
I’m 100% serious. Show me.
Feet pics thinks that if you lower the nozzles to 60 degrees the forward thrust and the thrust of the lift are equal
I literally don’t have to do anything to show you.
Look at the forward thrust vs lift thrust for 60 degrees of nozzle angle and tell me how on gods green earth its 50/50
45 is the actual 50/50 split and that’s what ive been telling you and him.
if the resultants are at 90 degrees if a vector is pointing 60 degrees offset to one its going to cause more force in that direction.
at 45 its equal
