Or you could just use the harrier 1 chart. Just an idea.
You have a Harrier 1 chart that shows bleed rate?
No?
I did not think so.
What?
If you angle the thrust down then not all of that thrust is pushing the plane forward. It is basic trigonometry.
No but it’s foolish to try compare it to the harrier 2 chart like you have been trying to do endlessly.
It’s extremely silly. The Harrier 1 chart shows the harrier 1 can sustain a full 6G at .8 Mach
The harrier 2 in that chart can’t even achieve .8 Mach in level flight.
Now you keep talking about the harrier 2 as being mediocre without taking 1 possible consideration into mind - what happens if a harrier 2 is using the normal lift thrust rating? What does the turn rate increase by then.
Seeing as the sustains G value on the harrier 1 was increased by over 1 in some cases. What’s to say seeing as they share the same engine and layout that the harrier 2 doesn’t achieve the same.
At 350 knots for example the Harrier 1 goes from sustaining 3.5G to sustaining 4.5G
What if the harrier 2 increased proportionally like the harrier 1?
3.8 to 4.8 at 350 knots if the added sustained G is the same as the harrier 1
So the harrier 2 goes from being mediocre to sustaining within 1 D/S with a clean F-16. Keep in minds that’s with all pylons and a gun pack. Vs the F-16 being clean, but only for this 1 given speed.
Regardless: We do not have any data to confirm the performance metrics of the Harrier 2 regarding normal lift thrust or CTVC
It’s such basic trigonometry that he got the angle wrong
Still at 45 degrees of nozzle angle the thrust going down and the thrust pointing aft are equal
What angle is wrong?
You said at 60 degrees the vertical and horizontal components are equal
Its actually equal at 45 degrees and at 60 more thrust is going down then aft
Read it again and read it slower. Where did I say that?
No, he said that at 60° the horizontal component of the thrust will be .50 of the total
The cosine of 60° is 0.5
Okay, but why are you only considering the horizontal component of thrust for an aircraft that is also in a turn?
Yes, literally what I just said.
cos60° = 0.5
sin60° = 0.866
U multiply those by the total thrust and u get the horizontal and vertical components respectively. They don’t have to sum the total thrust. Not even if u use 45° do they sum the total thrust, even if the components are equal
10,000lb of horizontal thrust is what proportion of 20,000lb? You can use your calculator for this problem.
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Right but the relationship doesn’t play out like that IRL
As he is implying that basic trigonometry is enough to determine all the factors in the calculations of vectored thrust propulsion.
Its not that simple - FAR from it. The relationship between the Net thrust and the component of Gross thrust added to the overall thrust changes.
So instead of 60 degrees creating a point of equilibrium it is actually 45 degrees.
You didn’t even use that initially lmao you used the gross thrust figures and decided to divide it in half for a given 60 degrees of nozzle angle.
I corrected you and said no its actually 45 0j and an equal split at 45 degrees and now you tried to backtrack to make it seem like you didn’t make this mistake.
I’m not holding that against you its just as simple as you tried to simplify something that basically can not be simplified.
there we go again
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Well, I suppose he was trying to simplify but yes the component of thrust acting vertically would also be pushing the aircraft into the turn.
There was no mistake. I used dynamic thrust values instead of static values.
Congratulations, you still used the incorrect nozzle angle to represent a 50/50 power split.
Regardless it’s not that big a deal.