MHB Formula for calculating DPS (damage per second)

  • Thread starter Thread starter El Demente
  • Start date Start date
  • Tags Tags
    Formula Per
El Demente
Messages
2
Reaction score
0
I am calculating damage per second (DPS) for a character in a video game.

This character strikes twice every 4th attack. (Doublestrike)

This extra strike does 50% of your attack damage.

This extra strike, along with all other strikes, can crit, meaning meaning they do twice the damage. My character can buy an infinity edge making crits do 2.5x the damage.

If my chance to crit (CC) is at 50%, my attack damage (AD) is 100, and my attack speed (AS) is 2 attacks per second, then what is my DPS with and without the infinity edge? (Assuming that I have attacked at least 4 times to proc the doublestrike)

What is the formula? (I need the formula because I am creating a dynamic spreadsheet)
 
Mathematics news on Phys.org
El Demente said:
I am calculating damage per second (DPS) for a character in a video game.

This character strikes twice every 4th attack. (Doublestrike)

This extra strike does 50% of your attack damage.

This extra strike, along with all other strikes, can crit, meaning meaning they do twice the damage. My character can buy an infinity edge making crits do 2.5x the damage.

If my chance to crit (CC) is at 50%, my attack damage (AD) is 100, and my attack speed (AS) is 2 attacks per second, then what is my DPS with and without the infinity edge? (Assuming that I have attacked at least 4 times to proc the doublestrike)

What is the formula? (I need the formula because I am creating a dynamic spreadsheet)

Hey El Demente! Welcome to MHB! ;)

Suppose your character attacks 4 times then the regular damage is:
$$4 \times AD + 50\% \times AD = 4.5 AD$$
The damage per attack is then:
$$\frac{4.5}{4} \times AD \cdot \frac{\text{damage}}{\text{attack}}$$

Multiply by attack speed to find the base damage per second. Let's call that $DPS_{\text{base}}$:
$$DPS_{\text{base}}= \frac{4.5}{4} AD \cdot \frac{\text{damage}}{\text{attack}} \times AS \cdot \frac{\text{attack}}{\text{second}}
= \frac{4.5}{4} \times AD \times AS \cdot \frac{\text{damage}}{\text{second}}$$

Since there is 50% chance on a critical hit, the expected damage is:
$$DPS_{\text{regular}} = 50\% \times DPS_{\text{base}}+ 50\% \times (2 \times DPS_{\text{base}}) = 1.5 \times DPS_{\text{base}}$$
With an infinity edge this becomes:
$$DPS_{\text{infinity edge}} = 50\% \times DPS_{\text{base}}+ 50\% \times (2.5 \times DPS_{\text{base}}) = 1.75 \times DPS_{\text{base}}$$

Summing it up:
\begin{array}{|l|c|c|} \hline
\text{DPS} & \text{Formula} & \text{Current Value}\\
\hline
DPS_{\text{base}} & \frac{4.5}{4} \times AD \times AS & 225\\
DPS_{\text{regular}} & 1.5 \times DPS_{\text{base}} & 337.5\\
DPS_{\text{infinity edge}} & 1.75 \times DPS_{\text{base}} & 393.75\\
\hline
\end{array}

Have fun! :cool:
 
Oh my god that is the perfect answer. Thank you so much. I input it into my spreadsheet and everything by the way! Kudo's and cheers! I knew I could wrap my head around it if someone broke it down for me I just got frustrated trying it forever myself on this one cause I knew I didn't quite have it.
 
Last edited:
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top