Question

Does geometrical base volume multiplied by geometrical density multiplier (or v multiplied by d) make sense?

Why

I'm trying to learn a bit of basic mathematics with a goal of applying it to my physics self-teach learning, with hope of better understanding the nature of geometry and gravity.

Formula

Cube object's base geometrical volume number (v) multiplied by geometrical density multiplier (d) so formula: v multiply by d

Picture

See my attached screenshot picture I took of 3-D objects I made - titled, "3D Conceptual Representation of Geometrical Objects and Density Multiplier". It was made with a 3-D software by myself, for conceptual illustration.

A) 4 * 1 = 4 (A simple cube with centre.)

B) 4 * 2 = 8 (A “tesseracted” cube with centre.)

C) 4 * 3 = 12 (A more denser object, also with centre as well.)

Similar to object's surface subdivision but different, only inside with a centre.

If we measure the mass of an object and get m

and measure the volume of the object and get V

you can get the density by dividing m/V

This will be expressed in units of M/L^3, such as kg/m^3

or g/cm^3

Then, when you find a similar object with similar contents, i.e. stone, water or iron,

you can estimate its mass simply by measuring its volume..

Measuring volume is much easier than measuring mass. You don't need any weight.

Just take the width x height x depth and you have the volume.

The mass of the Earth, the Sun, other planets or stars is not measured directly by applying a balance, but measured indirectly (which introduces some errors, possibly large ones).

17 hours ago, tylers100 said:

I'm trying to learn a bit of basic mathematics with a goal of applying it to my physics self-teach learning, with hope of better understanding the nature of geometry and gravity.

Check topic "unit cancellation"

"Unit cancellation, also known as dimensional analysis, is a problem-solving method used in various fields like physics, chemistry, and engineering. It involves using conversion factors to change units of measurement while ensuring the original value remains unchanged. This technique relies on the principle that multiplying any expression by 1 (represented by a conversion factor) does not alter its value, and it helps in converting units by canceling out unwanted units to arrive at the desired unit. "

https://en.wikipedia.org/wiki/Conversion_of_units

17 hours ago, tylers100 said:

Does geometrical base volume multiplied by geometrical density multiplier (or v multiplied by d) make sense?

Depends on what “geometrical base volume” and “geometrical density multiplier” mean.

Volume has a definition, as does density. Density is p= m/V, so pV is mass, by definition, so it had better make sense.

@swansont

Geometrical Base Volume and Geometrical Density Multiplier

I combined geometrical notion with some physical things in order to better approximate these, but I'm not sure if that is a right approach to do so because in the science of physics or an aspect of Nature itself (i.e. physics) might doesn't in fact have an actual geometry as being part of within but can be approximated and defined, I think. After all, the geometry is a fundamental concept constructed by humans to better approximate some of physics in Nature.

Reference link: https://en.wikipedia.org/wiki/Geometry

I made an error with the original post, see:

Error

I made an error in counting amount of density on the original post (was "geometrical base volume") in each of cubic objects; a, b, and c - was 4 when it should be 6 as baseline density number. The actual number is 6 density which counts 6 divided parts within cubic object A, 12 for B, and 18 for C. While A is baseline for tesseraction, B and C are tesseracted subdivision - with centre of geometry considered.

Tesseracting Construction with Geometry Notion

I think I have chosen the tesseraction model from which could theoretically tesseracting subdivision and mathematically construct objects to better understand the nature of gravity with geometry considered; these gravitational interactions and behaviours.

Newton's Law of Universal Gravitation

I'm still trying to learn and understand the Newton's Law of Universal Gravitation equation as stated on the following website:

Reference link: https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation

But my cognition and comprehension level is not there yet, so hence self-teach with basics of mathematics and physics for me for time being.

Formulas and Definitions

mass: m = p x V^3 = amount of matter

cubic volume: V^3 or if given m and p without V^3 then V^3 = m / p = amount of space that matter takes up

density: p = m / V^3 = amount of space that matter packed in

If formulas and definitions stated above are right, then:

Example 1 (Object 'A' with baseline for tesseraction)

mass: 6 p x 1^3 V = 6 m

cubic volume: 6 m / 6 p = 1 V^3

density: 6 m / 1^3 V = 6 p

Example 2 (increments with the density only while volume remain unchanged - Check the attached picture in the original post for reference - titled, "3D Conceptual Representation of Geometrical Objects and Density Multiplier")

a) 6 p x 1^3 V = 6 m

b) 12 p x 1^3 V = 12 m

c) 18 p x 1^3 V = 18 m

Example 3 (increments with the density and volume in a gradual way - different and hypothetical objects not referred in the picture or prior example but use similar mathematical approach)

d) 6 p x 1^3 V = 6 m

e) 12 p x 2^3 V = 96 m

f) 18 p x 3^3 V = 486 m

After learning a bit and if formulas and definitions are right as stated above, I'm a bit surprised and amazed to see results in the just above Example 3 section - how small adjustments to density and volume in a theoretical and gradual tesseracting construction of object(s) could yield such mass number because greater mass could logically means greater gravitational attraction or a possible change in gravitational interaction(s) or behaviour(s) between objects.

Point

@Sensei

Checking and ensure I get some of basic mathematics right in order to apply it to my physics self-teaching to make a progression with better understanding the gravity physics. I'll look into pre-algebra, algebra, and conversion of units at a later time.

Another Question

If cubic volume (V^3) is not known but still have a given of mass (m) and density (p), then m / p to obtain volumetric number. But what about a situation when a volume is not cubic one and volumetric un-measureable at that?

To elaborate my question's point:

Imagine two cubic objects being together forming a rectangle unit but its volumetric measurable is not yet known but for the sake of point let's assume and say; 2x, 1y, and 1z then would that make V^3 not usable for measurement because V^3 refers to a cubic in an uniform way meaning one of dimensional axes such as x has extra length (e.g. 2x) whereas y and z doesn't. Could that mean I have to use this mathematical approach: 2x x 1y x 1z = 2 V?

27 minutes ago, tylers100 said:

mass: m = p x V^3 = amount of matter

cubic volume: V^3 or if given m and p without V^3 then V^3 = m / p = amount of space that matter takes up

density: p = m / V^3 = amount of space that matter packed in

Volume is already “cubic”; the dimensions of V are L^3 (L is length)

27 minutes ago, tylers100 said:

If formulas and definitions stated above are right, then:

Example 1 (Object 'A' with baseline for tesseraction)

mass: 6 p x 1^3 V = 6 m

cubic volume: 6 m / 6 p = 1 V^3

density: 6 m / 1^3 V = 6 p

Your equations are inconsistent

3 hours ago, swansont said:

Volume is already “cubic”; the dimensions of V are L^3 (L is length)

Your equations are inconsistent

I have identified '1 V^3' as an error so here is corrected version:

volume: 6 m / 6 p = 1^3 V

The 'Example 1 (Object 'A' with baseline for tesseraction)' refers to a situation when 3 scenarios are considered;

1. If mass amount is absent, then given density and volume if available: p x V = m

2. If volume is absent, then given mass and density if available: m / p = V

3. If density is absent, then given mass and volume if available: m / V = p

1 hour ago, tylers100 said:

1. If mass amount is absent, then given density and volume if available: p x V = m

2. If volume is absent, then given mass and density if available: m / p = V

3. If density is absent, then given mass and volume if available: m / V = p

Yes, this is how density, mass and volume relate to each other. It’s really just one equation, which can be rearranged via basic algebra

Would it be workable if I adjust the mass formula to following, to make it nearly universal across spatial dimensions or is mass exclusive to length ^ 3 only?

Mass formula: ( density * ( length ^ spatial dimension ) ) mass

6 hours ago, tylers100 said:

Mass formula: ( density * ( length ^ spatial dimension ) ) mass

Your algebra is inconsistent.
That parses as mass².

It's simple, given P=density, M=mass, and V=volume
P=M/V or M=V*P or V=M/P