I often see it said that given a scheme $X$ and a graded sheaf of $\mathcal{O}_X$-algebras $\mathcal{S}$ the cone Spec $S$ has a natural $\mathbb{A}^1$-action on it induced from the grading on $S$.
But the grading on $\mathcal{S}$ is discrete ($\mathbb{Z}$-graded), whereas the action of $\mathbb{A}^1$ is continuous. What is this natural morphism $\mathbb{A}^1 \times \text{Spec } S \to \text{Spec } S$? I cannot see it.
Looking locally on $X$, we can assume $X=\operatorname{Spec} A$ and we can consider $\mathcal{S}$ as simply a graded $A$-algebra $B=\bigoplus_{n\in\mathbb{N}} B_n$, and $\operatorname{Spec} \mathcal{S}$ is just $\operatorname{Spec} B$. To give a map $\mathbb{A}^1\times \operatorname{Spec} \mathcal{S}\to \operatorname{Spec} \mathcal{S}$ then, we just have to give an $A$-algebra homomorphism $\varphi:B\to B[t]$. This homomorphism is simply the homomorphism that sends each $b\in B_n$ to $t^nb$.
For a simple example, take $A$ to be a field and $B=A[x_1,\dots,x_n]$ to be a polynomial ring with its usual grading. Then $\operatorname{Spec} B$ is just $\mathbb{A}^n$, and our action $\mathbb{A}^1\times\mathbb{A}^n\to \mathbb{A}^n$ is dual to the homomorphism $A[x_1,\dots,x_n]\to A[x_1,\dots,x_n,t]$ which sends $x_i$ to $tx_i$ for each $i$ (since each $x_i$ has degree $1$). Geometrically, this means the action sends $(t,x_1,\dots,x_n)\to (tx_1,\dots,tx_n)$. That is, it is just the usual action of $\mathbb{A}^1$ on $\mathbb{A}^n$ by scalar multiplication.
(Note that it is necessary for $\mathcal{S}$ to be $\mathbb{N}$-graded rather than $\mathbb{Z}$-graded here. If it is only $\mathbb{Z}$-graded, our homomorphism $\varphi$ would land in $B[t,t^{-1}]$ instead and we would get just a map $(\mathbb{A}^1\setminus\{0\})\times \operatorname{Spec} \mathcal{S}\to \operatorname{Spec} \mathcal{S}$, which is an action of $\mathbb{G}_m$ rather than $\mathbb{A}^1$.)