Leps 2.0 - step-by-step user guide. Nonlinear shear strength envelopes for rockfills Hi Folks, Here is a simple step-by-step user guide for the updated non-linear shear criterion for rockfills (https://lnkd.in/gA66TWiw). This is Version 0. As the criterion evolves, this guide will be updated. Practicality is the aim, and user feedback is welcome. Enjoy, and I hope it serves the purpose. practical and user-friendly 😊 !

I wrote a blog post about why you shouldn't look at a completed map and think "that's so obvious, why bother...". The value of mapping is in the making and the activities that need to happen to get to the stage where your map looks simple and self-evident Link in the comments

Repercussion of relative time on Datum due to absence of velocity model The fixed coordinates lose their physical position, causing: • epoch-dependent positional discrepancies, • false deformations, • incompatibility of datasets, and • potential legal disputes in cadastral surveys. In short: “A datum without velocity is a datum that decays with time.”

Day 16: Local Variables  A variable defined within a block or method or constructor is called a local variable.  The Local variable is created at the time of declaration and destroyed when the function completed its execution. The scope of local variables exists only within the block in which they are declared. We first need to initialize a local variable before using it within its scope. Example: This example show how a local variable is declared and used inside the main method and it can not be used outside of it. #JavaBasics #LocalVariables #JavaWithHanningtone

Problem: 189. Rotate Array Approach We need to rotate the array nums to the right by k steps. That means every element at index i should move to (i + k) % n, where n is the size of the array. Key Idea: Reverse Technique Instead of shifting elements one by one (which is inefficient), we can use the reverse method: Since rotating n times gives the same array, we only need k = k % n. Reverse operations: Reverse the first n - k elements. Reverse the last k elements. Reverse the entire array. This ensures that the last k elements come to the front while maintaining the correct order. Example Input: nums = [1,2,3,4,5,6,7], k = 3 Steps: Reverse first 4 → [4,3,2,1,5,6,7] Reverse last 3 → [4,3,2,1,7,6,5] Reverse whole → [5,6,7,1,2,3,4] Output: [5,6,7,1,2,3,4] ✅ Complexity Analysis Time Complexity: Each reversal takes O(n). Total = O(n). Space Complexity: In-place reversal, no extra array used. Total = O(1). ✅ This is the optimal solution with O(n) time and O(1) space. #DSA #DAY3

Cramer’s Rule gives an elegant and explicit formula for solving square systems of linear equations with a unique solution. It works well for small systems and for conceptual understanding, but for large or numerically sensitive systems other methods are preferred.

Seismic Activity: Beware! Don’t be spooked by potential seismic activity. Be prepared with CSS V-Wrap™. Seismic activity can creep up when you least expect it, but STRUCTURAL TECHNOLOGIES’ CSS V-Wrap™ system is a proven strengthening system engineered for flexibility, strength, and minimal disruption to help structures stand strong even against the scariest shakes. Featured Project: To keep U.S. 50 (Lincoln Highway) safe from potential threats, CSS V-Wrap™ was used to meet modern codes and reinforce this critical structure. Beware the risk. Be prepared with CSS V-Wrap™. Read the full story: https://lnkd.in/ef97btFh

Var is function-scoped and hoisted, allowing redeclaration. Let and const are block-scoped, but let allows reassignment while const does not (though objects/arrays can be mutated). Use const by default, let when needed, avoid var.

🚀 Day 13 – 90 Days of Code 🔍 Problem: Container With Most Water You're given an array height[] where each element represents the height of a vertical line at position i. Your goal is to find two lines that, together with the x-axis, form a container holding the most water (i.e., with the maximum area). 💡 Intuition: To form a container between two lines: The width is the distance between the two indices The height is the minimum of the two lines So, the area = min(height[i], height[j]) × (j - i) Now we want to maximize this area. ✅ Approach: We use the two-pointer technique for optimal efficiency: Initialize two pointers: left = 0 right = n - 1 While left < right: Compute area: area = min(height[left], height[right]) × (right - left) Update maxArea if this area is larger Move the pointer pointing to the shorter line inward (Since the limiting height is the smaller one) Continue until left >= right Return the maximum area found 🧮 Complexity: Time: O(n) — single pass from both ends Space: O(1) — no extra space used

Generators: Functions that yield values, saving memory. Decorators: Modify function behavior. Context Managers: For resources that need setup/teardown (e.g., files). Regular Expressions (Regex): import re. For pattern matching in strings.