# Tutoring math, you encounter grade. The tutor discusses its definition and why it might be surprising.

Grade is defined as 100%*(vertical/horizontal). In the above diagram, it would be as follows:

grade=100%*(rise/run).

By itself, rise/run is called slope.

Therefore,

grade=100%*slope

What follows is a distinction that, to me, is important and interesting:

grade is not

rise/distance traveled

since, of course, you can’t drive along the horizontal course of a hill; rather, you can only drive on its surface.

At level, grade and (rise/distance traveled) are both zero. They remain virtually the same even at 20% grade, when (rise/distance traveled) is 19.6%. As the grade increases, however, they differ dramatically: at 100% grade, (rise/distance traveled) is 70.7%.

My interest in the difference between grade and (rise/distance traveled) is philosophical: why base a value on an indirect measurement (horizontal distance), when a direct measurement (distance traveled) is available?

In math, we use slope, of course; however, it’s usually in a context where actual measurements aren’t used. Rather, it’s just on paper.

Source:

engineeringtoolbox.com

connect.ubc.ca

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring math, calculator features are always topical. The tutor mentions the ENG function on the Casio fx-991ES PLUS C.

Back in my Halloween, 2015 post, I explain the general idea of the ENG function – that engineers and technicians often like to refer to measurements in “thousand” groupings. For instance, 1249W is 1.249kW; 0.067A is 67mA.

Example: convert 0.78A to mA

1. Key 0.78 = (you must press = for this to work)
2. Press ENG: hopefully you see 780×10-3

Example: convert 1249W to kW.

Solution: to go up to the bigger unit, use SHIFT ENG, like so:

1. Key 1249 = (once again, you must press =)
2. Press SHIFT then ENG. Hopefully you see 1.249×103

HTH:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor gives background along with a basic explanation of how a high-pass filter works.

A high-pass filter will send along high frequency signals but block low frequencies. It can do so because the impedance, Xc, of a capacitor of capacitance C, at frequency f, is

Xc=1/(2πfC)

At very high frequency, therefore, Xc≈0

The total impedance, Z, of a resistor R and capacitor C in series is given by

Z=(R2 + Xc2)0.5

Therefore, with input voltage V, and instantaneous current i, Ohm’s Law gives

V=i(R2 + Xc2)0.5

The voltage across the resitor is

VR=iR

Therefore, the ratio of the voltage across the resistor to the source is

VR/V = iR/i(R2 + Xc2)0.5 = R/(R2 + Xc2)0.5

As mentioned earlier, for high frequency, Xc≈0, giving

VR/V ≈ R/(R2)0.5 = 1

With a high-pass filter, the maximum output voltage for a high frequency signal equals the input voltage.

Convention says that the critical frequency, fc, is that at which VR/V = 1/(2)0.5 = 0.707, which occurs when Xc = R:

VR/V = R/(R2 + Xc2)0.5 = R/(R2 + R2)0.5 = R/(2R2)0.5

which leads to

VR/V = 1/20.5 = 0.707

Therefore, a high-pass filter will pass any frequency higher than the critical frequency fc, where fc is calculated from

Xc = R

1/(2πfcC) = R

1 = 2πfcCR

1/(2πCR) = fc

By that reasoning, a 10kΩ resistor, in series with a 12pF (picoFarad) capacitor, placed in series, should produce a high-pass filter with critical frequency fc = 1.3MHz. The output would be read across the resistor R.

Source:

www.electronics-tutorials.ws

ecee.colorado.edu

Serway, Raymond A. Physics for Scientists and Engineers with modern physics. Toronto: Saunders College Publishing, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor tells how to access the built-in εο constant on the Sharp el-520w.

The Sharp el-520w has 52 built-in constants relating to physics, chemistry, etc.

Here’s how to call up εο, which has value 8.85×10-12:

1. Press the CNST key, and you’ll be asked which of constants (01-52) you require.
2. Key in 13. You’ll see 8.854187817×10-12 appear.

Source:

Sharp Scientific Calculator Model EL-520W Operation Manual.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor examines the idea that internal combustion engines are more efficient in cold weather.

An upper limit for efficiency of an internal combustion engine is

eff = (Tcombust – Tsurrounding)/Tcombust

where

Tcombust is the temp of the combustion cylinder

Tcombust, Tsurrounding both in degrees Kelvin (Celsius + 273).

Let’s imagine a diesel engine, whose average internal cylinder temperature might be around 1600°C. Then at outdoor temp of 25°C (293K) we have

eff = (1873 – 298)/1873 = 84.1%

Likewise, at outdoor temp -25°C we have

eff = (1873 – 248)/1873 = 86.8%

The 2.7% increase in efficiency at -25°C vs 25°C may be noticeable to an operator.

Source:

www.ncert.nic.in

White, J. Edmund. Physical Chemistry: College Outline Series. New York: Harcourt Brace Jovanovich, 1987.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor continues about a toy repair, with some reinforcement ideas.

While we were repairing the toy a few nights ago (see my previous post), my father-in-law suggested that, after the first repair cured, a second application should be considered around the outside. Such reinforcement, he commented, would give the repair its best chance of holding.

I considered his counsel from an engineering point of view: how much extra strength could we anticipate from application of J-B Weld around the outside of the repair site?

Let’s imagine the shear force to be straight forward. The strength of a reinforcement can be, generally, proportional to its left-right length multiplied by its height, then by the square of its forward length. Assuming the J-B Weld works as an integral piece after drying, I imagined an outside application along each side plane. The application would be about 20 times the height of the original shear, then its same forward length, but only about 1/30 of its left-right length. Compared with the first repair reuniting the two sheared surfaces, the reinforcement strength per side might be 20(1/30) or 2/3. Both sides together could offer reinforcement strength of 2(2/3)=4/3 or 1.33 times the strength of the original repair, more than doubling its shear resistance.

With these numbers in mind, I took my father-in-law’s advice and made the reinforcement application about 24 hours ago. The repair should be ready right now.

I’ll be sharing more about this fascinating toy repair:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor explains his recent understanding of a function he’s wondered about.

I’ve noticed the ENG function on more than one calculator, but have never used it. I’ve always assumed it means “engineering”; since I’m not one, it makes sense that I’m unfamiliar with it.

Yesterday my curiosity finally focused on this mysterious ENG function. You access it by SHIFT ÷ on the Casio fx-260solar. If you’re in COMP mode (I haven’t tried it with other modes), it seems to change the entered number to the highest power of 103 for which the number will be > 1. Examples:

0.056 SHIFT ÷ gives 56×10-3

0.000362 SHIFT ÷ gives 362×10-6

12037059.1 SHIFT ÷ gives 12.0370591×106

While I’m not an engineer, this notation is familiar to me. I know that in electronics, it’s common to refer to 0.056A as 56mA. Similarly, 0.000178A will likely by referred to as 178µA, also known as 178×10-6A. 12400000Ω would likely be referred to as 12.4MΩ (M=Mega=106).

I have more to say about the ENG function:)

Source:

Casio fx-260solar operation manual. London: Casio Electronics Co., Ltd.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.