Geography of earth

Can i see the calculations to prove this? Let me guess… I wouldn’t understand them

Did @Geo Lover 7 see this thread?

Trigonometry said no........ so no
I have a PhD...

Is the earth flat?

I had to simplify it a bit for those who don't have a PhD but this is the basics:
Consider the generalized form of the Jacobian elliptic functions in the context of a doubly periodic lattice:
sn(u,k)2+cn(u,k)2=1anddn(u,k)2+k2⋅sn(u,k)2=1,\mathrm{sn}(u, k)^2 + \mathrm{cn}(u, k)^2 = 1 \quad \text{and} \quad \mathrm{dn}(u, k)^2 + k^2 \cdot \mathrm{sn}(u, k)^2 = 1,sn(u,k)2+cn(u,k)2=1anddn(u,k)2+k2⋅sn(u,k)2=1,
where sn(u,k)\mathrm{sn}(u, k)sn(u,k), cn(u,k)\mathrm{cn}(u, k)cn(u,k), and dn(u,k)\mathrm{dn}(u, k)dn(u,k) are solutions to the differential equation
(dsn(u,k)du)2=(1−sn(u,k)2)(1−k2sn(u,k)2).\left( \frac{d\mathrm{sn}(u, k)}{du} \right)^2 = (1 - \mathrm{sn}(u, k)^2)(1 - k^2 \mathrm{sn}(u, k)^2).(dudsn(u,k)​)2=(1−sn(u,k)2)(1−k2sn(u,k)2).
Now, combining this with the Fourier series expansion for complex exponential periodicities:
f(x)=∑n=−∞∞aneinωx,wherean=1T∫0Tf(x)e−inωx dx,f(x) = \sum_{n=-\infty}^\infty a_n e^{i n \omega x}, \quad \text{where} \quad a_n = \frac{1}{T} \int_0^T f(x) e^{-i n \omega x} \, dx,f(x)=n=−∞∑∞​an​einωx,wherean​=T1​∫0T​f(x)e−inωxdx,
we can represent non-trivial periodic solutions in terms of modular transformations τ=ω2ω1\tau = \frac{\omega_2}{\omega_1}τ=ω1​ω2​​, where ω1\omega_1ω1​ and ω2\omega_2ω2​ are the half-periods of the lattice satisfying ω1τ−ω2=2mπi\omega_1 \tau - \omega_2 = 2m\pi iω1​τ−ω2​=2mπi.
For hyperbolic analogs, consider the identity:
cosh⁡2(x)−sinh⁡2(x)=1,\cosh^2(x) - \sinh^2(x) = 1,cosh2(x)−sinh2(x)=1,
which directly corresponds to the imaginary projection of elliptic integrals:
∫0ϕdθ1−k2sin⁡2(θ)=u,where u=am−1(ϕ,k).\int_0^{\phi} \frac{d\theta}{\sqrt{1 - k^2 \sin^2(\theta)}} = u, \quad \text{where } u = \mathrm{am}^{-1}(\phi, k).∫0ϕ​1−k2sin2(θ)​dθ​=u,where u=am−1(ϕ,k).
Finally, invoking Ramanujan's Master Theorem, the qqq-series expansion of the Dedekind eta function:
η(τ)=eπiτ/12∏n=1∞(1−e2πinτ),\eta(\tau) = e^{\pi i \tau / 12} \prod_{n=1}^\infty (1 - e^{2 \pi i n \tau}),η(τ)=eπiτ/12n=1∏∞​(1−e2πinτ),
where τ=ω2ω1\tau = \frac{\omega_2}{\omega_1}τ=ω1​ω2​​ connects elliptic integrals with modular forms, providing a transcendental bridge between the realms of trigonometric and elliptic phenomena.

oh how mesmerizing..

Well at least now you know that the earth cannot be a pyramid!

it can 100% be a pyramid. don't try to apply your made up concepts like 'math'(whatever tf that is) here.

it can 100% be a pyramid. don't try to apply your made up concepts like 'math'(whatever tf that is) here.

Educate yourselves y'all @sdfj

Videos can't beat maths

Made up????? Here's the proof that the earth is flat - GCSE trigonometry...
Begin by considering the Earth's surface as a 2D Euclidean plane in which distances d between two points (x1,y1) and (x2, y2) are calculated using the standard Euclidean metric:
d= sqrt((x2−x1)² +(y2−y1)²)
The sum of the angles (where A, B and C are each points on the earth):
α+β+γ=180∘
Trigonometric identities for the sine rule, valid only in flat geometry, are then applied:
alpha/sin(a) = beta/sin(b) = gamma/sin(c)
Using the formula for the curvature: K=1/L where K is the curvature of the earth and L is the longest distance between two points on the earth.
As L →∞, K→0
If K ~ 0, the curvature of the Earth is approximately equal to zero - therefore there are small bumps and hills on the earth as K is not equal to exactly zero but it doesn't curve! Therefore, the earth is flat!
@DerDracologe The flat earth proof is much more easy to understand...

Where are you getting these from 😭

Videos can't beat maths

I have a PhD (and a brain)

Did you mean to reply anonymously?

Accident
I guess people with PhDs aren’t quite perfect (almost, but not quite… )