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Welcome to 2021, and may 2020 stay (far) behind us!
Some fun things to occupy your time:
- Find the center of a circle
- Make some quiche
- Enumerate permutations of things
- Have some laughs here, here, here, here, or here.
I’m working on an Arduino project. Yes, I know I complained about them before, but that’s before I found out about the Espressif ESP32 and it’s smaller cousin, the ESP12 / ESP8266. These have built-in WiFi and Bluetooth, and have all sorts of cool features.
So, if you’ve visited this site in the last couple of years, you may have noticed that in the upper-right corner of the page, there is a widget called “Temperature at Casa de Parr”. This runs on a Raspberry Pi Zero W (like the Gate Controller). There is a DS18B20 temperature probe attached to the Pi, which is in a weather-proof case, and a second one connected via a wire, which sits just below the water level inside the pool.
At the time, the Pi was WAY overkill for this type of application, but for about $30 and some coding, I had a WiFi-accessible thermometer – very cool.
Well, I’ve been working on re-creating this using an ESP12, which is about $4, and I can use the same DS18B20 temperature probes.
Although the DS18B20 temperature probes are very consistent (precise), they are not very accurate, and therefore there is a need to store a calibration factor – basically an offset that you can add or subtract to the measured temperature, which may be two or three or even four degrees off, in order to calibrate it to the “real” temperature.
Of course, this assumes that the error is a constant at all temperature ranges – we could talk about two-point (linear-gradient) or three-point (curved) calibration, or even use an n-point bezier curve, which would be much more accurate, but for my needs, a constant is probably sufficient.
Unlike the Pi, which uses an SD card as its file system, the ESP can’t read or write any files unless you attach an SD card reader and write some code to read and write the files. Instead, they are designed to have a single, static program that gets stored in flash memory, which loads and runs automatically at boot. In addition, it has 512 bytes of EEPROM that’s accessible to the user, to store things like state or settings – 1/2 KB isn’t very much storage.
So to store my calibration factor, I need to write a float to EEPROM, and the method for writing to EEPROM is to write one byte at a time. A float is 4 bytes, and therefore I have to perform 4 write operations to 4 separate EEPROM addresses.
In order to accomplish this, I need to access my 4-byte float as a byte array. The easiest way to do this is to create a casted pointer, and I thought the concept behind this was interesting enough that you, dear reader, might be interested in it as well.
float f; char *c=(char *)&f; for(int i=0;i<4;i++) EEPROM.write(iBaseAddress+i,c[i]);
The “char” (or byte) data type is one byte – normally, you would use an array of char as a string:
char myString[]="Hi There!";
At this point, you can access the individual bytes of myString as elements in the array:
myString[3]=='T'; //true
In addition, I can create a char pointer to my string, and access the pointer as if it was my original string:
char *c=&myString; //& means "Address of"
char *c=myString; //The compiler knows what you're trying to accomplish
c[3]=='T'; //true
Using the address operator (&) we can assign the pointer c to the address of our string &myString. However, if you just assign a pointer to a string, most compilers know what you’re trying to accomplish, and just directly assigning it results in the same thing – the pointer c holds the address of myString.
If we try to do this with our float, we get a type conversion error:
float f=3.14;
char *c=f; //Error: Can't convert float to char
char *c=&f; //Error: Can't convert float to char
However, if we cast our float as a char, we are telling the compiler to treat the variable f (a 4-byte float) as if it was a char. Using casting with the address operator returns a pointer to a char, which really points to our float:
char *c=(char *)&f;
Now, c[0] through c[3] are the four bytes of our float.
I wrote this out within my code, and I got to thinking about it – I don’t do very much c coding these days, and I should probably double check. I found NO GOOD answers, and the top answer seemed to be to use a union:
union {
float f;
char c[4];
} fl;
fl.f=3.14;
for(int i=0;i<4;i++)
EEPROM.write(iBaseAddress+i,fl.c[i]);
This works, and I could certainly create a union type:
union fltype {
float f;
char c[4];
}
union fltype fl;
This is certainly more readable, but the only time I need to read a float as a byte array is when reading or writing to EEPROM. Using type casting just seems like a cleaner solution.
Well, I didn’t (If you said ‘no’ then response.write “either”) !
Here is an example that uses JavaScript to keep a color picker and a text input in sync with each other:
<script>
function sync(s,t) {
var a=document.getElementById(s);
var b=document.getElementById(t);
if (a.value!=b.value) b.value=a.value;
}
</script>
<input id="PickColor" type="COLOR" value="#207cff" onchange="sync('PickColor','HexColor');" />
<input id="HexColor" type="TEXT" value="#207cff" onchange="sync('HexColor','PickColor');" />
|
|
Here is the Color picker without the JavaScript:
<input name="WebColor" type="COLOR" value="#207cff" /> If you add this to a form, when the user submits the form, one of the form fields will be “WebColor”. |
If you have anything to do with technology, the chances are pretty good that you wasted at least part of the month of December patching, upgrading, or reconfiguring software that uses Log4J.
Now that the dust has settled, we can look at the vulnerabilities that were announced in December, what factors led to the problem, and why it should never have occurred.
It’s easy to play armchair quarterback by offering early criticism, so I felt that it was important to withhold any commentary until a majority of the facts came to light.
On October 21, 2021, in New Mexico, on a set for the movie “Rust”, a live round was discharged from a firearm, resulted in the fatal shooting of the cinematographer and injury to the director.
As a result of the incident, key people involved in handling the gun have been scrutinized by the media:
In addition, the shooting incident has sparked a debate about the general use of firearms as props in TV and movies.
I want to state unequivocally that I have no background in the film and theater industry, but I do have a level of knowledge and experience with many different kinds of firearms that exceeds at least 99% of the general public.
The purpose of this post is to review the facts of the incident (current, as of this writing), provide my analysis and thoughts about what may have led up to the incident, as well as what could and should have been done to prevent it.
I needed a way to list all of the Windows certificate stores…
Google failed me, so here it is:
Microsoft Windows [Version 6.1.7601]
Copyright (c) 2009 Microsoft Corporation. All rights reserved.
C:\Windows\system32>certutil -enumstore
(CurrentUser: -user)
LocalMachine
(CurrentService: -service)
(Services: -service -service)
(Users: -user -user)
(CurrentUserGroupPolicy: -user -grouppolicy)
(LocalMachineGroupPolicy: -grouppolicy)
(LocalMachineEnterprise: -enterprise)
My
Root
Trust
CA
TrustedPublisher
Disallowed
AuthRoot
TrustedPeople
Homegroup Machine Certificates
SmartCardRoot
SPC
TrustedDevices
Windows Live ID Token Issuer
CertUtil: -enumstore command completed successfully.
If you’re looking for the store names listed in MMC, they are listed with a completely different name, because… Microsoft:
| CertUtil Storename | In “Certificates” MMC | Purpose |
| My | Personal | Certificates assigned to this user or machine |
| Root | Trusted Root Certification Authorities | Root CAs trusted by this machine – typically this isn’t used very often |
| Trust | Enterprise Trust | Active Directory and other CAs related to management and authentication |
| CA | Intermediate Certification Authorities | Intermediate CAs trusted by this machine – typically this is not used. The server should serve out an intermediate that is downloaded on the fly, and must chain to a root CA in “Third-Party Root Certification Authorities” |
| AuthRoot | Third-Party Root Certification Authorities | Public trust providers such as DigiCert / GeoTrust or Thawte |
To list all of the certificates within a store:
C:\Windows\system32>certutil -store authroot authroot ================ Certificate 0 ================ Serial Number: 7777062726a9b17c Issuer: CN=AffirmTrust Commercial, O=AffirmTrust, C=US NotBefore: 1/29/2010 8:06 AM NotAfter: 12/31/2030 8:06 AM Subject: CN=AffirmTrust Commercial, O=AffirmTrust, C=US Signature matches Public Key Root Certificate: Subject matches Issuer Template: Cert Hash(sha1): f9 b5 b6 32 45 5f 9c be ec 57 5f 80 dc e9 6e 2c c7 b2 78 b7 No key provider information Cannot find the certificate and private key for decryption. ================ Certificate 1 ================ Serial Number: 600197b746a7eab4b49ad64b2ff790fb Issuer: CN=thawte Primary Root CA - G3, OU=(c) 2008 thawte, Inc. - For authorized use only, OU=Certification Services Division, O=thawte, Inc., C=US NotBefore: 4/1/2008 6:00 PM NotAfter: 12/1/2037 5:59 PM Subject: CN=thawte Primary Root CA - G3, OU=(c) 2008 thawte, Inc. - For authorized use only, OU=Certification Services Division, O=thawte, Inc., C=US Signature matches Public Key Root Certificate: Subject matches Issuer Template: Cert Hash(sha1): f1 8b 53 8d 1b e9 03 b6 a6 f0 56 43 5b 17 15 89 ca f3 6b f2 No key provider information Cannot find the certificate and private key for decryption. . . . . . ================ Certificate 52 ================ Serial Number: 01 Issuer: CN=AddTrust External CA Root, OU=AddTrust External TTP Network, O=AddTrust AB, C=SE NotBefore: 5/30/2000 4:48 AM NotAfter: 5/30/2020 4:48 AM Subject: CN=AddTrust External CA Root, OU=AddTrust External TTP Network, O=AddTrust AB, C=SE Signature matches Public Key Root Certificate: Subject matches Issuer Template: Cert Hash(sha1): 02 fa f3 e2 91 43 54 68 60 78 57 69 4d f5 e4 5b 68 85 18 68 No key provider information Cannot find the certificate and private key for decryption. CertUtil: -store command completed successfully.
And there you go, kids… always remember to use your powers for good and not evil.
This morning, July 20, 2021, Blue Origin’s New Shepard rocket launched in to space for its first 11-minute passenger flight.
Congrats to Bezos and team!
HOWEVER, the commentator on one of the science channels made this statement: “The rocket needs to go high enough to reach orbit, even briefly”
A popular misconception is that orbit and outer space are the same thing – actually, I’ve written about this previously. However, I would expect a science commentator on a science-related channel to know the difference!
Just to recap:
Even hundreds of thousands of miles away from the surface, any object is still within the Earth’s gravitational influence. So, if you fire a rocket straight up (perpendicular to the Earth’s surface), even hundreds or even thousands of miles in to space, eventually, the rocket will run out of fuel, and the Earth’s gravity will pull it back down.
In orbit, however, an object’s velocity parallel to the Earth’s surface creates centrifugal force which balances the pull gravity.
E. A rocket launched perpendicular to the Earth’s surface eventually runs out of fuel and is pulled back down by gravity. F. To achieve orbit, a rocket must attain both height and velocity parallel to the surface.
My understanding of New Shepard is that it will basically travel straight up, give the passengers a few minutes of weightlessness, and then return to Earth.
Although gravity will be acting on the spaceship and its passengers during the descent, the passengers will experience weightlessness (which is different than outer space and also different than orbit) because gravity acts on both nearly-equally. The state or sensation of weightlessness occurs when there are no net forces acting on the passengers, which is only true inside the ship, and only relative to the ship itself.
Although the exact height of the New Shepard’s mission isn’t listed, it is expected to go “well above” the Kármán Line. So if we were to guess 65 miles, we would be in the right ball park.
This means that New Shepard’s velocity at 65 miles above the Earth’s surface will be zero, after its rockets turn off, and gravity bleeds away all of its upward velocity. At that exact point, with gravity acting constantly, it will begin to accelerate back down to Earth. Until the ship hits turbulence in the atmosphere, or they fire their re-entry rocket, the passengers will continue to fall at an ever faster rate toward Earth, pulled by gravity, yet experiencing weightlessness within the ship.
In contrast, to attain orbit at 65 miles above the Earth’s surface, the ship would have to be travelling almost 17,780 miles per hour parallel to the Earth’s surface in order to have enough centrifugal force to exactly counter the force of gravity.
How do we go about calculating this? Glad you asked…
Force of gravity:
F = m ⋅ g
- m = mass of the object
- g = acceleration due to gravity
Centrifugal force (same as centripetal force, but in the opposite direction):
F = m ⋅ v2 r
- m = mass of the object
- v = linear velocity
- r = radius of the orbit (from the center of the Earth)
In orbit, the net force is zero, so:
m ⋅ g = m ⋅ v2 r We quickly see that mass is irrelevant, and we can solve for v:
v = √ g ⋅ r
If we measure gravity and orbital radius in feet, we get orbital velocity in feet per second. To simplify things, we can use 78,545 miles per hour2 as the acceleration due to gravity:
v = √ 78545 ⋅ r
The radius of the Earth is about 3960 miles. To find orbital velocity, we have to have the radius from the center of the Earth, which include’s the Earth’s radius plus the height above the surface:
v = √ 78545 ⋅ ( 3960 + r )
If we plug in 65 miles, we get about 17,780 miles per hour required for orbit.
So again, good luck to Bezos and team – although you won’t be “in orbit”, you will definitely be in outer space.
Previously, I’ve written about gun-related movie myths in “Movie Myths: Guns – Part 1“, and in there, I described in detail how guns work – both automatics and revolvers.
Since that time, I’ve thought about making a state diagram that helps explain some of the common gun-related continuity errors that you regularly see in movies and TV.
The result is the diagram that you will see in this article, along with a higher-resolution PNG and PDF that you can download for free, which would make a nice wall decoration for any gun enthusiast.
Download the PDF here, or the PNG here.
Read on for more information about state diagrams and gun-related movie continuity errors…
I bought a wood lathe recently, and the first time I tried to center some wood, I went through the painful process of remembering how to find the center of a circle. This, despite having had three years of mechanical drawing classes.
Starting with a pencil and straight edge:
The center of the circle is at the intersection of the vertical (E) and horizontal (H) center lines.
Let’s call the distance from the corner of the straight edge to its mark S.
Let’s call the points where the horizontal center line meets the circle’s edges H1 (left) and H2 (right), and the point where the vertical center line meets the bottom V.
If we start at M and walk clockwise around the circle, we would pass through the points as follows: M (start) → H2 → V → H1 → M (back to start).
If the radius is 1, then the diameter is 2, and the length of MH2 is 1.41. We initially selected S to be about 3/4 of the diameter, or about 1.5.
The circle’s radius, R, is 1/2 the length of MV or H1H2. The length of MH2 would be √ R2 + R2 , and the same is true for H2V, VH1, and H1M.
If we assume that the radius R is “1 unit” (the actual length is irrelevant), then √ 12 + 12 = √ 2 , or about 1.41.
If the radius R = “1 unit” then the diameter = 2 ⋅ R = “2 units”.
By selecting S such that S is 3/4 of the diameter, 3/4 ⋅ 2 = 1.5. Since S>1.41, and we make two consecutive S-length marks around the circle, we know for a fact that 2 ⋅ S is greater than the length of MH2V.
By doing this twice, once clockwise, and once counterclockwise, we get lines C and D. We can call their intersection point T
MAD = MBC = 2⋅S. Point T is located at the intersection of lines C and D. Triangles ΔMAT and ΔMBT are congruent. MT bisects AB.
⋅
It’s easy to prove that this forms two congruent triangles: ΔMAT and ΔMBT. If the line MT is the base of both, then because they are congruent, the height of both are the same, which means that MT perfectly bisects AB).
Since this line MT also bisects the equilateral triangle ΔMAB which is inscribed within the circle, that line must pass through the center of the circle as well.
Carrying MT out to the edge of the circle gives us point V.
By drawing a line from V of length S to the edge, we get line F, which crosses line A. We can call this intersection point U.
Points U and W are at the intersections of A and F, and B and G respectively. MUV is equilateral because MU=UV. MUV is congruent to MWV.
If we knew where the center of the circle (point O) was located, we could make the congruent triangle argument, that ΔMUO is congruent to ΔVUO, and just as with the vertical center line bisecting ΔMAB, UO would perfectly bisect the larger equilateral triangle ΔMUV.
On the other side, we draw line G of length S, which crosses line B at point W. Same congruency argument: ΔMWO is congruent to ΔVWO, and WO bisects equilateral triangle ΔMWV.
Because MU = VU we know that ΔMUV is equilateral, and the same for ΔMWV. Further, because of the congruency argument, we know that MU = MW and VU = VW, and therefore ΔMUV is congruent to ΔMWV.
If line UO bisects triangle ΔMUV, then it must intersect MV exactly at its center. Same for WO and triangle ΔMWV.
Therefore, WU passes through the center of both triangles at point O, where WU intersects MV.
And therefore, point O must be at the center of the circle, because VO = MO, and we know that MV passes through the center of the circle.
Note that this trick ONLY works if the the mark on the ruler (distance S) is > 1.41 times the radius.