average with respect to time vs with respect to distance

mean average with respect to time vs with respect to distance

Notation

I am writing 2*(1second)^2 with a shorthand of 2 second ^ 2 that is why I am writing meter not meters and second not seconds.  I am doing this because it became very confusing writing out the units differently for plural and singular form such as 2 seconds but 1 second when the units were put to powers other than 1.

x = displacement

t = time

m = mass

f = force

a = acceleration

v = velocity

P = momentum

KE = Kinetic Energy

Int[y, g(y)] = the anti derivative of g(y) with respect to y

dg(y)/dy = the derivative of g(y) with respect to y

Dyg(y) = the derivative of g(y) with respect to y

Example 1

given x(t=0 seconds) = 0 meter

given v(t=0 seconds) = 0 meter second ^ -1

given a(t) = 1 meter second ^ -2

v(t) = t*(1 meter second ^ -2)

x(t) = 0.5 meter second ^ -2 * t^2

t^2 = x / (0.5 meters * seconds ^ -2 ) 

t^2 =  2*x / ( 1 meter per second ^ 2 )

t^2'= (2 meter^-1 second ^2) * x

t(x) = (2^0.5)*(1 meter ^ -0.5 second ^ 1) *(x^0.5)

v(x) = t(x) * (1 meter second ^ -2)

v(x) = (2^0.5)*(1 meter ^ -0.5 second ^ 1) *(x^0.5) * (1 meter second ^ -2)

v(x) = (2^0.5)*(1 meter ^ 0.5 second ^ -1) *(x^0.5)

antiderivative of v(x) with respect to x shall be called Int[x, v(x)]

Dx(x^N) = N*x^(N-1)

Int(x, x^N) = (1/[N+1])*x^(N+1) + K0

Dx(1/[N+1])*x^(N+1) = (N+1)/(N+1)*x^(N+1-1)

Int[x, v(x)] = (1/1.5)*(x^1.5)*(2^0.5)*(1 meter ^ 0.5 second ^ -1) + K1

Int[x, v(x)] = (2/3)*(x^1.5)*(2^0.5)*(1 meter ^ 0.5 second ^ -1) + K1

Int[x, v(x)] = (x^1.5)*(2^1.5)*(1 meter ^ 0.5 second ^ -1) / 3 + K1

x(t) = 0.5 meter second ^ -2 * t^2

x (1 second) = 0.5 meter second ^ -2 * (1 second) ^ 2 = 0.5 meter

Int[x, v(x= 0.5 meter)] = (0.5 meter^1.5)*(2^1.5)*(1 meter ^ 0.5 second ^ -1) / 3 + K1

0.5 ^ 1.5 * 2 ^ 1.5 = 1

Int[x, v(x=0.5 meter)] = ( 1 meter ^2 second ^ -1 ) / 3 + K1

Int[x, v(x=0 meter)] = ( 0 meter ^2 second ^ -1 ) / 3 + K1

Average velocity with respect to displacement from a displacement of 0 meters to a position of 0.5 meter =

= ( Int[x, v(x= 0.5 meter)] - Int[x, v(x=0 meter)] ) / ( 0.5 meter - 0 meter) =

= ( 1 meter ^ 2 / 3 ) / 0.5 meter =

= 2 meter / 3

antiderivative of v(t) with respect to t shall be called Int[t, v(t)]

v(t) = t*(1 meter second ^ -2)

Int[t, v(t)] = 0.5 meter second ^ -2 * t^2 + K2

Int[t, v(t=1 second) = 0.5 meter second ^ -2 * (1 second )^2 + K2

Int[t, v(t=1 second) = 0.5 meter + K2

Int[t, v(t=0 second) = 0 meter + K2

Average velocity with respect to time from a time of 0 second to a time of 0.5 second =

= ( Int[x, v(t= 1 second)] - Int[x, v(t=0 second)] ) / ( 1 second - 0 second) =

= (0.5 meter-0meter) / (1 second-0second) = 0.5 meter second ^ -1

given mass = 1 kilogram

KE(x) = 0.5 * 1 kilogram * [v(x)]^2

v(x) = (2^0.5)*(1 meter ^ 0.5 second ^ -1) *(x^0.5)

KE(x) = 1 kilogram meter second ^ -2 * x

F(x) = dKE/dx = 1 Kilogram meter second ^ -2

P(t) = 1 kilogram * v(t)

v(t) = t*(1 meter second ^ -2)

P(t) = t*(1 kilogram meter second ^ -2)

F(t) = dP(t)/dt = 1 kilogram meter second ^ -2

The average velocity with respect to time is not the same as the average velocity with respect to displacement for an object that starts stationary and experiences constant acceleration when averaged over a distance interval that is the same distance as it would have traveled in the time interval in which it's average velocity with respect to time is calculated.  However the average force it would experience over such an interval would be the same if either averaged by time or by displacement because the force is the same constant amount in either case.  Such would hold true for a falling object experiencing a constant force having displacement, time and velocity measured as zero at the initial starting condition of the fall.

Example 2

given a(t) = t^2 * 1 meter second ^ -4

given v(t=0 second) = 0 meter second ^ -1

given m = 1 kilogram

given x(t=0 second) = 0 meter

v(t) = 4*t^3 * 1 meter second ^ -4

x(t) = 12*t^4* 1 meter second ^ -4

[t(x)] ^ 4 = x / 12 meter second ^ -4

[t(x)] ^ 4 = x*(1/12)*1 meter ^ -1 second ^ 4

t(x) = x^(0.25) * (1/12)^0.25 * 1 meter^-0.25 second ^1

v(x) = 4 * 1 meter second ^ -4 * [t(x)]^3

v(x) = 4*1 meter second^-4*[x^(0.25)*(1/12)^0.25*1 meter^-0.25 second ^1]^3

v(x) = 4 * (1/12)^0.75 * x^0.75 * 1 meter ^0.25 second ^ -1

KE(x) = 0.5 * 1 kilogram * [v(x)]^2

KE(x) = 8*(1/12)^1.5*x^1.5*1 kilogram meter ^0.5 second ^ -2

Dx(x^1.5) = 1.5*x^0.5

dKE(x)/dx = 12*(1/12)^1.5*x^0.5*1 kilogram meter ^0.5 second ^ -2

f(x) = dKE(x)/dx

int[x, f(x)] = KE(x) + K1

x(t) = 12*t^4* 1 meter second ^ -4

x(t = 1 second) = 12 meter second ^ -4 * (1 second)^4 = 12 meter

KE(x) = 8*(1/12)^1.5*x^1.5*1 kilogram meter ^0.5 second ^ -2

(1/12)^1.5*12^1.5=1

KE(x = 12 meter) = 8*1 Kilogram meter ^2 second ^ -2

v(t) = 4*t^3 * 1 meter second ^ -4

KE(t = 1 second) = 0.5 * 1 kilogram * [4 meter second ^ -1] ^ 2

KE(t = 1 second) = 8 kilogram meter ^ 2 second ^ -2

average force with respect to displacement from 0 meter to 12 meter =

=(int[x, f(x = 12 meter)] - int[x, f(x = 0)])/(12 meter - 0 meter) =

= 8 kilogram meter ^ 2 second ^ -2 / 12 meter =

= (2/3) * 1 kilogram meter second ^ -2

v(t) = 4 * t^3 * 1 meter second ^ -4

P(t) = 1 Kilogram * v(t)

P(t) = 4 * t^3 * 1 Kilogram meter second ^ -4

P(t = 1 second) = 4 * (1 second)^3 * 1 Kilogram meter second ^ -4

P(t = 1 second)  = 4 kilogram meter second ^ -1

f(t) = dP(t)/dt = 12 * t^2 * 1 Kilogram meter second ^ -4

int[t, P(t)] = P(t) + K2

average force with respect to time from 0 second to 1 second =

=(int[t, f(t= 1 second)] - int[t, f(t = 0)])/(1 second - 0 second) =

=  4 kilogram meter second ^ -1 / 1 second =

=  4 kilogram meter second ^ -2

f(x) = 12*(1/12)^1.5*x^0.5*1 kilogram meter ^0.5 second ^ -2

f(x=12meter)=12*(1/12)^1.5*(12meter)^0.5 *1kilogram meter ^0.5 second ^ -2

12*(1/12)^0.5*12^0.5=12

f(x = 12meter) = 12 kilogram meter ^ 2 second ^ -2

f(t) = dP(t)/dt = 12 * t^2 * 1 Kilogram meter second ^ -4

f(t = 1 second) = 12 kilogram meter second ^ -2

In this example the instantaneous force as a function of time is equal to the instantaneous force as a function of displacement at the displacement value that would occur for a given time value.  But the average force as a function of time is not the same as the average force as a function of displacement for the displacement interval that matches with the time interval.

Copyright Carl Janssen 2021 December 16

Estimating the gas laws based on collissions

Estimating the gas laws based on collissions

Starting with a simple one dimensional force calculation problem 

An object is traveling back and forth between two walls at a constant speed called travel speed except during collissions.  During collissions the object slows down to 0 then speeds up to it's original speed but in the opposite direction. The walls are parrellel to each other and a distance called travel length apart except during collissions where the wall hit by the object temporarily deforms then returns to it's original shape or is temporarily displaced then returns to it's original position after each collission ends.  The object moves in a path perpendicular to the walls.  The time the object is not in a collission is called travel time and the time the object is in a collission is called collision time.  The length a wall is temporarily displaced in the direction parrelel to the objects path of travel during a collission is called collission length.  The wall is moved or deformed during collission and the object is treated like a point mass with no deformation to make the math simpler.  

Mass refers to the mass of the object in one dimensional force calculations below.  Force below refers to a scalar being the absolute value of the force vectors magnitude and not to the direction of the force vector.  That is I am not counting forces going in opposite directions as having opposite values in the calculations below.  If forces were treated as vectors and no absolute value was taken there would be a zero average force when an equal number of complete collissions in each opposite direction occured.

The mean force the object would experience with respect to time using impulse = change in momentum = force * time

The absolute value of the change in momentum in a single collission is twice the absolute value of the objects momentum in this case because it switches to the same momentum in the opposite direction.

mean force with respect to time during travel = 0

mean force with respect to time during collission = 

= 2*mass*(travel speed) / (collission time)

mean force with respect to time = 

= 2*mass*(travel speed) / ([travel time] + [collission time])

(travel time) = (travel length) / (travel speed)

limit of mean force with respect to time as [(collision time) / (travel time)] approaches 0 is =

= 2*mass*(travel speed) / (travel time) = 

= 2*mass*([travel speed]^2) / (travel length)

The mean force the object would experience with respect to distance using work = force * distance

Work occurs during a collission to slow the object down to zero speed then again a second time to return it to original speed.  So the work done in a single collission is twice the traveling kinetic energy of the object in this case.

mean force with respect to distance during travel = 0

mean force with respect to distance during collission =

= 2*0.5*mass*([travel speed]^2) / (collission length)

mean force with respect to distance =

= 2*0.5*mass*([travel speed]^2) / ([collission length]+[travel length])

limit of mean force with respect to distance as [(collision length) / (travel length)] approaches 0 is =

= 2*0.5*mass*([travel speed]^2) / ([travel length])

= mass*([travel speed]^2) / ([travel length])

Start Side Note

the mean force with respect to distance is half the mean force with respect to time when you assume collission length and collision time are both insignificantly small compared to travel length and travel time.

See Article

http://teachingthenarrowway.blogspot.com/2021/12/average-with-respect-to-time-vs-with.html

http://web.archive.org/web/*/http://teachingthenarrowway.blogspot.com/2021/12/average-with-respect-to-time-vs-with.html

End Side Note

Now Let's instead assume there is a three dimensional object of a cube of length L and instead multiple particles collide into the walls in any one of six directions or three orthogonal / perpendicular directions and there opposites each of those six directions being perpendicular to one of the six faces of the cubes.  In this case we will assume the particles do not move any other direction than those six directions to make the math easier so that in each case it will work mathematically similar to the one dimensional case above.  That is we will assume the particles will only collide with the faces of the cube moving in a direction perpendicular to the face they hit making the collission calculatable as a one dimensional problem in each case.  Let us also assume the particles do not collide into each other or do not effect each other if they were to collide into another particle but "phase through" each other or act as though having "no clipping" with respect to each other but only collide with cube faces/walls or only experience collission forces through cube faces/walls and not other particles.  Additionally we shall assume all particles have the same speed except during collissions even though in reality their would be a distribution of different speeds at most temperatures people normally experience.

Temperature = Average Kinetic Energy per molecue

(Travel Speed)^2 = Kinetic Energy per molecue / mass per molecue =

= Temperature / (mass per molecue)

Limit as [(collision time) / (travel time)] approaches 0 of (Average Force per Molecue) with respect to time  = 

= 2* (mass per molecue) *([travel speed]^2) / (L)

= 2*(mass per molecue)*[Temperature / (mass per molecue)] / L

= 2*Temperature/L

From now on I shall assume collission time is negligible and refer to force as the force calculated using the calculus limit above and as measured as on average with respect to time

Total force from all molecues = 

= Force per molecue * Density * Volume / (mass per molecue)

 = Force per mole * Density * Volume / (mass per mole)

= (2*Temperature/L) * Density * Volume / (mass per mole)

Surface Area of a cube = 6*(L^2)

Volume of a cube = L^3

Pressure = Total Force / Surface Area =

(2*Temperature/L)*Density*(L^3)/[(mass per mole)*(6*[L^2])]

Pressure = Temperature*Density/[3*(mass per mole)]

N = Density*Volume/(mass per mole)

PV=NrT

P*V = r*Temperature*Density*V/(mass per mole)

P = r * Temperature*Density/(mass per mole)

Temperature*Density/[3*(mass per mole)] = r * Temperature*Density/(mass per mole)

r = 1 / 3

I remember getting  a different answer of r = 2 / 3 when I calculated this in high school.  I decide to search for the phrase unitless gas constant online and see if anyone actually calculated a unitless value or if I am the only person to ever do so.  Apparently someone else also got 2/3 which is different than the 1/3 (or 1/6 because I calculated force two different ways based on average force as a function of length vs average force as a function of time) value I calculated this time but the same as I remember calculating in high school.

https://duckduckgo.com/?q=unitless+gas+constant&ia=about

https://en.m.wikipedia.org/wiki/Talk:Gas_constant#2%2F3_cut_-_ref_please

http://web.archive.org/web/20211215070129/https://en.m.wikipedia.org/wiki/Talk:Gas_constant#2%2F3_cut_-_ref_please

One might conclude that there should be different pressure depending on the shape of the container if it has a different volume to surface area ratio than that of a cube since the calculations were made using a cube shaped container but an assumption was made molecues do not experience force when they collide with each other only the container holding them in the calculations above.  If you remove the assumption of lack of molecular collission force except against the container walls the molecues would travel a much shorter distance between collissions.  You can imagine if there are N molecues in the container then imagine the container is made of N cubes,  the molecues would each inhabit a average volume of a cube shaped like the volume of the container / N and travel on average a length of the cube root of the (volume of the container divided by N) and have a average surface area of the cube shaped zone they move around in equal to 6 times the square of the cube root of the (volume of the container divided by N) you would end up with the same results as calculated above for the average pressure and force from these molecular collissions by partititioning a non cubed shape into many cubes.  

It might make more sense to assume the molecues typically inhabit a sphere shaped shape than a cube but a sphere which has a diameter equal to the length of a side of a cube has an equal surface area to volume ratio to that of a cube but would have a smaller surface area and volume than that cube requiring more spheres to fit in the same container if they each had a diameter equal to a side length of a cube as such it would result in a slightly different unitless constant number for R, especially if the molecues were considered to be able to move in more than six directions and a distribution of speeds rather than all moving at the same speed except when colliding.  

Pressure * Volume = Force/Area * Length/Area = Force * Length = Energy

Number of Moles * Temperature = Number of moles * Kinetic Energy per mole = Energy

Some number of Joules  = PV = rNT = r* some number of Joules

Some number of calories  = PV = rNT = r* some number of calories

r = some number without units

Even though r is technically unitless it is listed as a ratio of several units which cancel out each other to allow people to convert units from one type to another.

Surface Area / Volume for a cube of side length L

(6 L^2) / (L^3) = 6 / L

Surface Area / Volume for a Sphere 

(4*pi*r^2)/(4*pi*(r^3)/3) = 3 / r

(pi*D^2)/(0.5*pi*(D^3)/3) = 6 / D

Surface Area for a E*L by F*L by G*L rectangular solid

2*(EF+FG+EG)*(L^2)

Volume for a E*L by F*L by G*L rectangular prism

 E*F*G*(L^3)

Surface Area / Volume for a E*L by F*L by G*L rectangular prism

2*(EF+FG+EG)/(E*F*G*L)

Make a rectangular solid that is not a cube with the same surface area / volume as a cube

Let 2*(EF+FG+EG)/(E*F*G) = 6

When E = 2 and F = 1

2*(2+G+2G) / 2G = 6

12 G = 4 + 6G

6 G = 4

G = 2 / 3

2*(2+[2/3]+[4/3]) / [4/3] = 2*4 / [4/3] = 6

Let L = 1 inch

1 inches by 2 inches by 2/3 inch

volume = (4/3) inches^3

surface area = 2*([2*1]+[2*2/3]+[1*2/3]) inches^2 = 8 inches^2

8 inches^2 / (4/3) inches ^3 = 6 inches ^ -1

1 inch by 1 inch by 1 inch cube

6 inches ^ 2 / 1 inch ^ 3 = 6 inches ^ -1

Copyright Carl Janssen 2021

In some video games, noclip mode is a video game cheat command that prevents the first-person player character camera from being obstructed by other objects and permits the camera to move in any direction, allowing it to pass through such things as walls, props, and other players.

Noclipping can be used to cheat, avoid bugs (and help developers debug), find easter eggs, and view areas beyond a map's physical boundary.

http://web.archive.org/web/20170520115201/https://en.m.wikipedia.org/wiki/Noclip_mode

https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-8b/v/antiderivative-of-x-1

http://web.archive.org/web/20190515181746/https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-8b/v/antiderivative-of-x-1

https://socratic.org/questions/what-is-the-antiderivative-of-ln-x-4

http://web.archive.org/web/20211216034307/https://socratic.org/questions/what-is-the-antiderivative-of-ln-x-4

In the kinetic theory of gases, the mean free path of a particle, such as a molecule, is the average distance the particle travels between collisions with other moving particles. 

https://en.m.wikipedia.org/wiki/Mean_free_path

http://web.archive.org/web/20220629235820/https://en.m.wikipedia.org/wiki/Mean_free_path

https://en.m.wikipedia.org/wiki/Kinetic_theory_of_gases

http://web.archive.org/web/20220513084951/https://en.m.wikipedia.org/wiki/Kinetic_theory_of_gases

Why you should be able to skip high school geometry and go straight for trigonometry

 It was common in the late 20th century and early 21st century (AD/ACE) for high schools and colleges to teach pre algebra followed by alegebra then geometry, then trigonemtry (or Algebra 3 / Triginometry) then precalculus and finally a course equivelent to AB and BC AP calculus after that there are several calculus courses that all require a course equivelent to AB and BC calculus as prerequisites.

Pre algebra was an unnecesary course because it was a mixture of the math taught before Algebra in other math classes and Algebra that you would take a second time when you take the Algebra course.  Thus if you completed all the math classes that were prerequisites for Pre Algebra you could skip it and go straight to algebra and there would be nothing you would be taught in Pre Algebra that was not either taught in Algebra or the prerequisite courses for prealgebra that you would miss.

Precalculus was a combination of Algebra 3 trigonometry and AB calculus with the exception of the possible inclusion of matrixes and probability in Alegebra 3 and or Precalculus being included in one but not the other.   If matrixes and probability are included in both Algebra 3 Trigonometry and Precalculus then there is nothing someone would miss by skipping Precalculus and going straight to AB calculus from Algebra 3 Trigonometry.  If Matrixes and Probability were not included in both Algebra 3 Trigonometry and Precalculus then there is nothing someone would miss by skipping Precalculus and going straight to AB calculus from Algebra 3 Trigonometry.

The reason for being able to remove Pre Algebra and Pre Calculus courses and let someone skip them is overlapping and redundant teaching material with zero new teaching material added.  But there is a diffetent reason all together for the lack of necessity of the high school or college geometry course in that sequence in addition to any overlapping material.

The real life practical purpose of geometry is to measure distances or lengths of paths, distances between objects or lengths of objects, locations (coordinates) of objects, volumes of objects and cross sectional and surface areas of objects.  All that can be done using material usually taught in high school algebra, trigonometry and calculus classes without ever needing to take a high school geometry course.  If you understand caartesian coordinates taught in algebra class, polar coordinates taught in trigonometry and or calculus class and the meaning of the sin, cos, tan and cotangent functions and there inverses as well as what radians are and the pythagorean theoreom taught in trigonometry class you will be able to do every one of the practical purposes I have listed that you would be able to do taking a geometry class.  The only one of those things you might learn in calculus that you might not learn in trigonometry is polar coordinates which really should be taught in trigonometry before AB calculus because it is more of a trigonometry related thing.  Now there are some of those practical things I listed you might learn to do in calculus that you would not learn in high school trigonometry but you would not learn them in high school geometry either.  Examples would be learning how to calculate the surface area or cross sectional area or volume of certain shapes by integration or the distances of certain paths that are not a single straight line segment by integration.  Now there are some useful things involving geometry people might learn like names of shapes but people usually do not learn those in high school geometry but in other classes prior to high school geometry.  In Geometry class you might learn complicated arcane and obscure ways to measure those practical things I listed in very specific special cases that might not be taught in trigonometry but in trigonometry you will learn general wide reaching methods to do all those things you can do in geometry which will be easier to remember because you do not have to create a special case to figure out each thing but can simply figure out the coordinates of things using trig functions then use those coordinates to get the desired measurements.

* Start side note

For further reading about how to get trigonometric measurements by means of algebra with casrtesian and polar coordinates without geometry class you can read the following article in progress

Tangent of the average of two angles and other trigonometric identities derived from a combination of it and the pythagorean theoreom

http://web.archive.org/web/*/http://teachingthenarrowway.blogspot.com/2022/01/tangent-of-average-of-two-angles-and.html

http://teachingthenarrowway.blogspot.com/2022/01/tangent-of-average-of-two-angles-and.html

* End side note

But what about the philosophy of using axioms taught in high school geometry being useful to learn how to use logic to make practical decisions.  This so called philosophy taught in high school geometry is the most important reason it should be abolished except perhaps for those who wish to study how cultic thinking works.  High School Geometry is full of the bad kind of dogma that is presuppositions that one is not permitted to question, as opposed to stating presuppositions you use to come to a conclusion but acknowledging the possibility your presuppositions maybe wrong.  The starting point is there is a finite number of assumptions the student starts with called axioms and the claim is made that all geometric proofs in the geometry class can and will be taught either by using only those axioms or other proofs derived only through those axioms.  The student is given a list of those axioms at the beginning of the course and expected to derive all proofs assigned as homework only through those axioms or through other proofs they have derived eventually tracing back to a point of only through those starting axioms.  The fundamental problem with this is in reality assumptions are made to derive these proofs that are not in the list of initial axioms but the student is not permitted to admit that additional assumptions are required other than the axioms initially listed if you say another assumption is needed for the proof that is not listed in the initial list given at the beginning of the course that is classified as a thought crime and the problem is marked as wrong.  The type of erroneous methodology of proofs in high school geometry class influenced most participants (who mostly were not previously educated about cult psychology to develop a resistance to the undue influence) towards pseudomathematics, pseudoscience, pseudologic and magical thinking through the act of saying and or writing things in order to agree with the consensus of an authority figure even if those things are not true.  

In the Asch conformity experiment people were more likely to claim a untrue statement was true if someone else first claimed the same untrue statement was true.

The Asch conformity experiment was an experiment where volunteers were told a false answer about the length of a line segment then asked what length the line segment was.  It was found that when a unanimous group of people gave a false answer first people more frequently said the same measurement as the false answer they were previously told instead of the correct measurement than under other circumstances.

 Something similar to the Asch conformity experiment is replicated in geometry classes where the teachers assign what assumptions maybe used and students are not allowed to pick the assumptions they use for themselves and describe which assumptions they use and how and why they used them.  Even if the teachers claims about what assumptions are needed to prove a genuinely true statement is true are false by ommission or commission the students will (or more accurately have in the past implying others will in the future) more ftequently still give the same answer as the teacher than if they never heard the teachers claim.  False by ommission in this context would be where the teacher claims that such is the exact minimum list of assumptions required to validly prove a true claim when actually given the context on the list the teacher gives additional assumptions need to be added.  False by commission in this context would be to claim a assumption is needed to validly prove a true statement when it is not needed in the context of the other listed assumptions, that is if all the other assumptions on the list were included but that assumption was removed it would still be a valid proof of a true statement.  In addition a teacher may sometimes claim a false statement is true or a true statement is false and the students will more ftequently give the same answer as the teacher.  If the students did not first hear the teacher make a false claim or say a true claim is true based on logic that was not valid the students would be much less likely to make the same error when investigating if a statement is true independently in an ungraded environment without seeing someone else's example first.  Remember in geometry class people are not graded for right or wrong answers but they are graded for right or wrong reasoning.  

According to the way this type of geometry class is done the reasoning should be in agreement with the reasoning predefined in the curriculum before the student even started the class in order to not be marked as wrong or less than an A or 100% grade for a set of problems.  One of the fundamental problems is the assumption of minimum axioms that is the foundation for all reasoning in high school geometry classes of the type defined in this essay/article/manifesto is usually or maybe even always wrong based on a claim of being complete when more assumptions are actually needed in reality.  Say you claim the teacher or textbook is missing an assumption when they show how to do a geometric proof which was an assigned homework or test problem and that assumption is not in the axiom list or derived from the initial axiom list alone and you will get your problem marked as wrong.

If a list of five assumptions were given to reach a conclusion and you were asked if these five assumptions alone with nothing added or removed were valid to show the conclusion was true or untrue and nobody told you the answer beforehand and you were not given a hint what answer someone else came to and nobody else would see or grade your answer this would not be so similar to Asch conformity.  But in the case of geometry class the teacher starts by listing a certain number of assumptions and says they are sufficient.  According to the principle learned in the Asch conformity experiment by giving you the answer first people are more likely to verbally say the same answer they were first given by someone else even if the answer they were given is false.   Geometry classes of this sort do not start with the question of if those axioms are sufficient to prove the proofs but the statement that all the proofs you are required to prove in the class can be proved with those assumptions and no more excluding that which is derived from those assumptions alone.  Nobody in their right mind would believe no more assumptions will be needed based on independent thinking without someone else first telling them no more assumptions would be needed.  Those who are first told no more assumptions will be needed to prove the proofs in the book believe so (or have said they believe so) without even first reading the entire book more often than they should by which I mean more often than never.  Geometry classes of this sort are like religious instiuitions such as Churches or Mosques where they far too many members say they believe every word in a certain book such as a particular Bible or Quran or book of Mormon are true without ever having read the entire book.  Many self proclaimed Muslims can not read Arabic but claim every statement in the original Quran manuscript in Arabic is 100% true and many people who say they are Christian at Churches say every statement in the original Bible manuscripts of a particular Bible Canon are true without having even read an entire translation of it.  It is very different to have read a translation of the Bible and say you have found zero historical claims in it that you can confirm with great certainty to be false when correctly translated than to have never even read it and to claim every statement in it is true even statements you have not read.  Yet some geometry teachers will give their testimony to the axioms as being sufficient for all proofs in the assigned geometry book and far too many people will believe them without reading the entite book.  I am not saying to read an entire geometry book but do not give me your testimony of the sufficiency of the axioms to prove every proof in the book if you never read the entire book.

Copyright Carl Janssen 2021, 2022

https://en.m.wikipedia.org/wiki/Asch_conformity_experiments

http://web.archive.org/web/*/https://en.m.wikipedia.org/wiki/Asch_conformity_experiments

A 2003 effort (Meikle and Fleuriot) to formalize the Grundlagen with a computer, though, found that some of Hilbert's proofs appear to rely on diagrams and geometric intuition, and as such revealed some potential ambiguities and omissions in his definitions.[11]

[11] On page 334: "By formalizing the Grundlagen in Isabelle/Isar we showed that Hilbert's work glossed over subtle points of reasoning and relied heavily, in some cases, on diagrams which allowed implicit assumptions to be made. For this reason it can be argued that Hilbert interleaved his axioms with geometric intuition in order to prove many of his theorems."

http://web.archive.org/web/20210508113205/https://en.m.wikipedia.org/wiki/Hilbert%27s_axioms

Euclid's list of axioms in the Elements was not exhaustive, but represented the principles that seemed the most important. His proofs often invoke axiomatic notions that were not originally presented in his list of axioms.[23]

[23] Heath 1956, p. 62 (vol. I)

http://web.archive.org/web/20211209025416/https://en.wikipedia.org/wiki/Foundations_of_geometry

Absolute geometry is a geometry based on an axiom system for Euclidean geometry without the parallel postulate or any of its alternatives. Traditionally, this has meant using only the first four of Euclid's postulates, but since these are not sufficient as a basis of Euclidean geometry, other systems, such as Hilbert's axioms without the parallel axiom, are used.[1] The term was introduced by János Bolyai in 1832.[2] It is sometimes referred to as neutral geometry,[3] as it is neutral with respect to the parallel postulate.

[1] Faber 1983, pg. 131

[2] In "Appendix exhibiting the absolute science of space: independent of the truth or falsity of Euclid's Axiom XI (by no means previously decided)" (Faber 1983, pg. 161)

[3] Greenberg cites W. Prenowitz and M. Jordan (Greenberg, p. xvi) for having used the term neutral geometry to refer to that part of Euclidean geometry that does not depend on Euclid's parallel postulate. He says that the word absolute in absolute geometry misleadingly implies that all other geometries depend on it.

http://web.archive.org/web/20211209025836/https://en.m.wikipedia.org/wiki/Absolute_geometry