Look, I know the rocket travels up at an exponential rate.
Not arguing with that.
The fact the rocket burns fuel and loses mass in its flight has nothing to do with measuring the average velocity of the figures given.
Since you were earlier trying to brush off the fact the missile didn't vary far from vertical with an argument offering an example of using calculus to measure the area found under a triangle or trapezoid (which I naturally ignored as we are dealing with a curved trajectory in the instance), in order to find average velocity, what are you going to do now?
It is this simple.
The trajectory of the Hwasong -15 missile in November of 2017 was such that at an altitude of 250km, it varied from vertical of launch point 0 to a point no more than 50 km down range, more than likely near 30km.
Go ahead and apply your calculus to determine average velocity of that profile and state the measure.
I will tell you right now the result would not differ significantly from the one derived using the linear calculator provided.
In addition, an object under no propulsion at an altitude of 250km, experiencing g=9.08m/s2, will not, under any circumstance, gain an additional 4250km of altitude.
Have a great day.
Facepalm.
I wasn't talking about the shape of the flightpath, I was talking about the shape of the velocity - time graph. That, is, fire the rocket straight up (so there is only one component of velocity to worry about) and plot its velocity (y-axis of the graph) against time (x-axis of the graph).
To calculate distance travelled you need the area under the graph. That is the velocity multiplied by the time, for every infinitesimally small chunk of time. For a simple profile, it's easy maths. If velocity is constant over a set time, then it's just the area of a rectangle - the velocity multiplied by the time. If it's a linear acceleration from zero, it's now a straight sloping line - a triangular shape. So the distance travelled is 1/2 x base x height, or in other words, the starting velocity (zero) plus the end velocity all divided by two, multiplied by the time. That's what your average velocity equation was doing.
But if the velocity profile is a more complex shape, and because of the changing mass and hence variable acceleration it absolutely is (a progressively steeper up-sloping curve in this case) then to find the area under the graph you have to do the calculus I described.
Do you now understand why the horizontal component is completely irrelevant?
For the final time, I have attempted to maintain a pretty decent level of decorum in this thread.
Your little comments like "facepalm" are not appreciated, nor are they necessary.
Your continued objections that average velocity,
d=rt, and all the other things I have pointed out that do not happen to match the narrative of this fairy tale, and suddenly could not or do not matter is simply related to your recognition they do not match the narrative. If I travel 250km in 5 minutes and I do not GAF if I am traveling in a circle or if I am traveling in a straight line or if I am traveling in one direction and then suddenly veer off to the left or right, at the end of the trip I have averaged a rate of travel equivalent to 3000km/h over the course of that trip.
Period. End of sentence.
For purposes of this discussion, that rate of travel is equivalent to VELOCITY since we are discussing a scalar quantity.
The numbers you want to claim for the rocket do not fit the requirements.
So, every time you guys want to tag up and bury these facts, do not forget I will be here, pointing out the claimants have not provided the math.
In case you forgot, you (along with the rest) are the claimants.
Instead of providing the math (which I am the only one to do so far, truth be told) all you guys have done is say, "nuh uh."
Go ahead and post your figures for an average velocity of a projectile that travels 250km in 5 minutes.
Post your math demonstrating a ballistic object traveling at a velocity of 16,000 km/h, at an altitude of 250km, experiencing g= 9.08m/s
2 under no further propulsion, will gain an additional altitude of 4250 km to apogee.
It is that simple.