Sensitivity and Weight [Crazy Theory?]

[qBe]

I'll try to make this short and precise.

Anyone else feels that HighSensi + heavy mice (or mice with more friction / bad mouse feet)

= low sensi + light mice ?

this is the formula to use conversion

Mice 1 = 105 gm

Mice 2 = 70 gm

3.6(sensi)/105(weight) = 0.03428571428 , sensi per gm

70gm mice = deflac = 105-70 = 353.6- (35*0.03428571428) = 3.6-1.1999999998 = 2.4000000002 (~2.4)

so 3.6 sensitivity with Mice1 would feel same with 2.4 Sensitivity with Mice 2 when DPi of both is same (i.e 400)

tbh, high sensi with heavy mice feels better

[RugOp'81]

Yes, ..crazy theory !
2 different eDPI is not the same, nor feels the same.

[reg1s7]

I don't think that is a correct calculation. Aside from the weight of the mouse, you also have the weight of your hand up to the resting point (your elbow or your wrist). And the weight of your hand does not become heavier when you use a heavier mouse.

[cJ]

Not sure about the accuracy of the calculations, but I never considered the idea of compensating for heavy mice with high DPI (and vice-versa). Would make an interesting video... maybe right up 3kliksphilip's alley?

[Ray Benchmark]

That's not how it works.

[⎛ PlamAtom ⎞]

Let's look at the whole situation from a purely physical point of view, you have a mouse with 0.070 Kg and 0.105 Kg. Let's assume both mice have an eDPI of 1000, meaning they move 707 Pixels per 2.54 cm, we can ignore the coefficient of sliding friction for polyamide on another polymer, assuming, of course, that everywhere the mouse moves there is a mousepad underneath and no force is excerted in a angel above 8 degree.

Under consideration it becomes clear that both mice cover the same distance on the screen which is dependent on the force exerted on a mouse, this force is controlled by the human and probably comes from the muscle memory or is much more variable, like in the case of a fresh player. This is expressed by F = m*a - Since our mouse is not in free fall, we can neglect gravitation.

Let us now consider the following scenario. We have a player who has changed his mouse from 0.122 kg (Titanwolf Gauntlet) to 0.135 kg (Rival 710). He does not want to retrain his muscle memeory, which exerts a certain force F on the mouse via the player's body. Let's consider this in an environment where the player can move a maximum of 2.54 cm (1 Inch). So how can we calculate the difference between the force applied in newtons to see how much we need to adjust the eDPI? the acceleration a is thereby meter*s². To measure this we take a camera and place it above the mouse, now we load a CS map and ask the player to aim at a corner but not to click. The player does this several hundred times and from that we take the average. Some attempts where the user slips, flicks or there are extreme delays will not be counted.

(Example with values from one observation, i.e. values that are almost certainly incorrect... i mean F is more than 2)

The player moves m122g on average 2.54cm in 0.08 seconds
The player moves m135g on average 2.54cm in 0.11 seconds

F.m = mass*(meter*second²)
F.m122g = 0.122 Kg*(0.254 Meter *0.08² Second)
F.m135g = 0.135 Kg *(0.254 Meter*0.11² Second)

F.m122g = 1.983232×10−4 (0.0001983232) Newton
F.m135g = 4.14909×10−4 (0.00041490900000000004)Newton

Difference in F = 2.0920850409836 Newton

The easiest way now would be eDPI / F-Difference - Directly offset the force with the actual distance traveled on the screen.

* This is just an attempt of mine and not the sound use of formulas... I am a mathematician and not a physicist. Personally, I simply did not find a value for the coefficient of friction, for example, and therefore simply assumed that it can be ignored. I would be happy if you check the calculations and add comments. It can also be that, the muscle memory affects due to various factors so that you can not make an accurate calculation. Thank you.

[Ray Benchmark]

Originally posted by ⎛ PlamAtom ⎞:

Let's look at the whole situation from a purely physical point of view, you have a mouse with 0.070 Kg and 0.105 Kg. Let's assume both mice have an eDPI of 1000, meaning they move 707 Pixels per 2.54 cm, we can ignore the coefficient of sliding friction for polyamide on another polymer, assuming, of course, that everywhere the mouse moves there is a mousepad underneath and no force is excerted in a angel above 8 degree.

Under consideration it becomes clear that both mice cover the same distance on the screen which is dependent on the force exerted on a mouse, this force is controlled by the human and probably comes from the muscle memory or is much more variable, like in the case of a fresh player. This is expressed by F = m*a - Since our mouse is not in free fall, we can neglect gravitation.

Let us now consider the following scenario. We have a player who has changed his mouse from 0.122 kg (Titanwolf Gauntlet) to 0.135 kg (Rival 710). He does not want to retrain his muscle memeory, which exerts a certain force F on the mouse via the player's body. Let's consider this in an environment where the player can move a maximum of 2.54 cm (1 Inch). So how can we calculate the difference between the force applied in newtons to see how much we need to adjust the eDPI? the acceleration a is thereby meter*s². To measure this we take a camera and place it above the mouse, now we load a CS map and ask the player to aim at a corner but not to click. The player does this several hundred times and from that we take the average. Some attempts where the user slips, flicks or there are extreme delays will not be counted.

(Example with values from one observation, i.e. values that are almost certainly incorrect... i mean F is more than 2)

The player moves m122g on average 2.54cm in 0.08 seconds
The player moves m135g on average 2.54cm in 0.11 seconds

F.m = mass*(meter*second²)
F.m122g = 0.122 Kg*(0.254 Meter *0.08² Second)
F.m135g = 0.135 Kg *(0.254 Meter*0.11² Second)

F.m122g = 1.983232×10−4 (0.0001983232) Newton
F.m135g = 4.14909×10−4 (0.00041490900000000004)Newton

Difference in F = 2.0920850409836 Newton

The easiest way now would be eDPI / F-Difference - Directly offset the force with the actual distance traveled on the screen.

* This is just an attempt of mine and not the sound use of formulas... I am a mathematician and not a physicist. Personally, I simply did not find a value for the coefficient of friction, for example, and therefore simply assumed that it can be ignored. I would be happy if you check the calculations and add comments. It can also be that, the muscle memory affects due to various factors so that you can not make an accurate calculation. Thank you.

While this is very interesting and funny to read, it doesn't really apply to the situation because we talk about weight differences of mices on a level nobody can really tell the difference considering how far and fast he moves them.

What I mean by that is that we don't need any more measureable energy to move the mouse from a to b while it weights 70g or 105g. In our brain we move to mouse to a specific position, force applied for that isn't important because it's not noticable. That's like if you are holding 10 ants in your hand or 30. It is no weight difference and for you it's not harder to hold them.

I don't know if I explained it good enough.

Also the suggested and average eDPI of good player is around 900 no matter what mouse they use. Because the distance will always be the same no matter how much a mouse weights. It's just the force applied more, which again, isn't even noticeable really.

[reg1s7]

Originally posted by Ray Benchmark:

While this is very interesting and funny to read, it doesn't really apply to the situation because we talk about weight differences of mices on a level nobody can really tell the difference considering how far and fast he moves them.

What I mean by that is that we don't need any more measureable energy to move the mouse from a to b while it weights 70g or 105g. In our brain we move to mouse to a specific position, force applied for that isn't important because it's not noticable. That's like if you are holding 10 ants in your hand or 30. It is no weight difference and for you it's not harder to hold them.

I agree. Even if we take weight difference into account, the weight of our hand should make the difference in the weight of the mouse less significant.

Sure increasing/decreasing the sensitivity might be a sound advice, but multiplying it by 1.5 like OP suggested should be a step too far.

[Ray Benchmark]

Originally posted by EvanS_:

Originally posted by Ray Benchmark:

While this is very interesting and funny to read, it doesn't really apply to the situation because we talk about weight differences of mices on a level nobody can really tell the difference considering how far and fast he moves them.

What I mean by that is that we don't need any more measureable energy to move the mouse from a to b while it weights 70g or 105g. In our brain we move to mouse to a specific position, force applied for that isn't important because it's not noticable. That's like if you are holding 10 ants in your hand or 30. It is no weight difference and for you it's not harder to hold them.

I agree. Even if we take weight difference into account, the weight of our hand should make the difference in the weight of the mouse less significant.

Sure increasing/decreasing the sensitivity might be a sound advice, but multiplying it by 1.5 like OP suggested should be a step too far.

True, based on OPs idea we should also increase/decrease our sens depending on the weight of our arm/hand. And when we lose/gain weight we should change our sense based on that. Some people gain/lose multiple kg of weight on their arm alone.