### Re: Is Time Real?

Posted:

**February 26th, 2017, 10:29 am**In 60 seconds, I am thinking divide the extra circumference by 2pi, or just call it 6 in head math. So roughly .16 m.

Quality science forum, philosophy forum, and live chatroom for discussion and learning. All are welcome, beginners and experts alike.

http://sciencechatforum.com/

Page **12** of **12**

Posted: **February 26th, 2017, 10:29 am**

In 60 seconds, I am thinking divide the extra circumference by 2pi, or just call it 6 in head math. So roughly .16 m.

Posted: **February 26th, 2017, 12:08 pm**

So my question to you is, the ratio of the extra length to the circumference is 1:40,000x1000x1000 = 1:40,000,000,000 so why do you not take that ratio rather than 1/2Pi?

Posted: **February 26th, 2017, 4:35 pm**

OK firstly let me tell you, you are right. Despite the fact the extra length is only 1/40,000,000,000 times more than the circumference of the Earth, it will sit about 160 mm or about 6.3 inches above the surface of the Earth

And here is the reason:

Let C2 be the second circumference and C1 the first, and r2 and r1 be the radii corresponding to the two circumferences.

Then C2 – C1 = 2 x Pi x r2 – 2 x Pi x r1

Therefore rearranging the equation above:

r2 – r1 = (C2 – C1) ÷ 2Pi

or the difference between the radii = the difference between the circumferences ÷ 2 Pi - Always and forever more

So whether the circumference of the first sphere is a mere 1 cm or half an inch or the size of the sun, if the difference is 1 m, then that's all that matters.

And here is the reason:

Let C2 be the second circumference and C1 the first, and r2 and r1 be the radii corresponding to the two circumferences.

Then C2 – C1 = 2 x Pi x r2 – 2 x Pi x r1

Therefore rearranging the equation above:

r2 – r1 = (C2 – C1) ÷ 2Pi

or the difference between the radii = the difference between the circumferences ÷ 2 Pi - Always and forever more

So whether the circumference of the first sphere is a mere 1 cm or half an inch or the size of the sun, if the difference is 1 m, then that's all that matters.

Posted: **February 26th, 2017, 9:29 pm**

Hi Raj,

Well, 40,000,000 meters divided by 40,000,001 meters is going to be a long fraction of like .99999999999xxxx... etc. Then I shifted the number 0.9999999 left 2 digits to get 99.9 Millimeters. I should have shifted it 3 digits and gotten 999.9 millimeters. But anyway, after rounding off my incorrect answer (99.0 mm) I got 100 Millimeters, so I must be off by at least a factor of 10.

But even 100mm feels like too large a difference.. but I suck at guessing. I even suck at the Math unless I'm real careful. I'm tempted to calculate the real answer just to find out how badly I guessed at it.. and screwed up the meter to millimeter conversion to boot.

Ok.. I give.. I'll do the real calculation now and see what the answer is.. stand by..

Holy crap, the number is larger than what I thought it would be and my screwed up answer above isn't so far off, assuming I did the Real Math correctly just now (of course).

I won't spoil your test by submitting my Calculated answer. I'll PM you my real calculated answer, which still feels too large (and the procedure I used to derive said answer).

Oops.. you already gave the answer above. I didn't see the thread flowed onto another page already.

So I won't PM you my results and just show mine here anyway:

So my calculation was:

D1=40,000,000 / Pi (diameter sphere)

D2=40,000,001 / Pi (larger sphere)

D2-D1 = X (difference in Diameters)

X/2 = elevated difference on one side = .159154943 meters

or

159.154943 Millimeters

Dang, that's much larger than what I would have though it to be.

6.2659 inches? For just one more meter in circumference. That's mind blowing!

Best wishes,

Dave :^)

Well, 40,000,000 meters divided by 40,000,001 meters is going to be a long fraction of like .99999999999xxxx... etc. Then I shifted the number 0.9999999 left 2 digits to get 99.9 Millimeters. I should have shifted it 3 digits and gotten 999.9 millimeters. But anyway, after rounding off my incorrect answer (99.0 mm) I got 100 Millimeters, so I must be off by at least a factor of 10.

But even 100mm feels like too large a difference.. but I suck at guessing. I even suck at the Math unless I'm real careful. I'm tempted to calculate the real answer just to find out how badly I guessed at it.. and screwed up the meter to millimeter conversion to boot.

Ok.. I give.. I'll do the real calculation now and see what the answer is.. stand by..

Holy crap, the number is larger than what I thought it would be and my screwed up answer above isn't so far off, assuming I did the Real Math correctly just now (of course).

I won't spoil your test by submitting my Calculated answer. I'll PM you my real calculated answer, which still feels too large (and the procedure I used to derive said answer).

Oops.. you already gave the answer above. I didn't see the thread flowed onto another page already.

So I won't PM you my results and just show mine here anyway:

So my calculation was:

D1=40,000,000 / Pi (diameter sphere)

D2=40,000,001 / Pi (larger sphere)

D2-D1 = X (difference in Diameters)

X/2 = elevated difference on one side = .159154943 meters

or

159.154943 Millimeters

Dang, that's much larger than what I would have though it to be.

6.2659 inches? For just one more meter in circumference. That's mind blowing!

Best wishes,

Dave :^)

Posted: **February 27th, 2017, 12:12 am**

Hi Dave,

Yep it is mind boggling. But as I pointed out in the general algebraic solution this string will sit at around 159 mm (1/2pi) higher, whether the sphere is just 1 m in circumference and the string is 2 m long or whether the sphere is 1 million Kms in circumference and the string is still just 1 m longer.

It's the difference that counts not the circumference, if you want to know how much more the radius will be.

Yep it is mind boggling. But as I pointed out in the general algebraic solution this string will sit at around 159 mm (1/2pi) higher, whether the sphere is just 1 m in circumference and the string is 2 m long or whether the sphere is 1 million Kms in circumference and the string is still just 1 m longer.

It's the difference that counts not the circumference, if you want to know how much more the radius will be.

Posted: **February 27th, 2017, 2:15 pm**

Now this is for Dave alone, because he torments me by throwing meaningless numbers at me and claiming they unravel the mysteries of the Universe.

Here is a simple proof that 1 = 2

1. Assume that a = b

2. Multiply both sides of the equation by b, therefore: ab = b^{2}

3. Subtract a2 from both sides, therefore: ab – a^{2} = b^{2} – a^{2}

4. Factorise both sides, therefore a(b - a) = (b + a)(b – a).

5. Divide both sides by (b – a), therefore a = b + a

6. Since b = a, from (1), substituting in (5) above, we get a = a + a

7. Therefore: a = 2a

8. Dividing both sides by the common factor a, we get: 1 = 2

Ponder over this

Here is a simple proof that 1 = 2

1. Assume that a = b

2. Multiply both sides of the equation by b, therefore: ab = b

3. Subtract a2 from both sides, therefore: ab – a

4. Factorise both sides, therefore a(b - a) = (b + a)(b – a).

5. Divide both sides by (b – a), therefore a = b + a

6. Since b = a, from (1), substituting in (5) above, we get a = a + a

7. Therefore: a = 2a

8. Dividing both sides by the common factor a, we get: 1 = 2

Ponder over this

Posted: **March 1st, 2017, 4:13 am**

Hi Dave,

What's up? No response? :)

What's up? No response? :)

Posted: **March 1st, 2017, 11:26 am**

rajnz00 » February 27th, 2017, 6:15 pm wrote:7. Therefore: a = 2a

8. Dividing both sides by the common factor a, we get: 1 = 2

If a = 2a, then a = 0.

Therefore division by a is division by zero, which is not a legitimate operation, since contradictions can be derived from it.

Posted: **March 1st, 2017, 3:09 pm**

Positor » March 1st, 2017, 11:26 am wrote: "If a = 2a, then a = 0. "

If a = 2a then it doesn't follow that a = 0. a is legitimately any number.

However you're onto it when you say division by 0 is not allowed.

Where I have divided by 0 is in step (5).

b - a = 0, since they are both equal to each other.

Posted: **March 1st, 2017, 7:30 pm**

Hi Raj,

Sorry.. Monday I got called by my boss (I work from home usually) to work on a manual and had to bail. Yesterday, I had to go into work and solve a problem I couldn't solve from home. I left at 9:30 AM Tuesday (to work) and got home at 1:30 AM Wednesday. A 16 hour work day. Just got up two hours ago, for a call from work, then checked in here.

Again.. sorry I can't be on here as much as I like, but my job has my priority. But honestly, I would have solved such by plugging in a real number for A (given A=B) and worked it through and it would have failed somewhere along the procedure (step5?).

Anyway, I admittedly suck at Math. Now.. 45 years ago.. I was top of the class and aced Algebra. But I never use it in the real world. For the type of computer programming I do, Algebra is pretty rare. Even in Electronics it's rare.

Back in College this problem was submitted to the whole School. Given a month to work on it, I was the first to submit an answer Immediately. 60 people submitted answers over the next few weeks, including 10 professors. I was the only person with the correct answer.

Want to try it?

Years later, at a company that makes measuring equipment in electrically noisy environments, my Boss, with a PHD in Mathematics, submitted the ultimate algorithm using Linear Regression for filtering Noise from data gathered and digitized. He included a proof that it was unbeatable. It was a monster algorithm and took huge CPU and Memory Resources to run it.

I beat it with a statistical filter of my own design that took about 30 Bytes of Memory and about 20 lines of code, and run about 1000 times faster. He was furious.

But those were my glory days when I kicked butt against Experts in many fields. Today, I'm a has-been. My last Project (starting soon, before I can retire), requires me to do Voice Analysis and Speech Recognition. The Math on this one will be intense and insane. I have to do it on a small circuit board with a small CPU, little Memory, all self contained in a single stand-alone product. All my current Products Talk now, but this one has to Listen. I'm using a generic DSP. Needless to say.. it is scaring the crap out of me.. but I always succeed when I put my mind to a problem (so far.. lol).

A minor mention.. we have drifted radically from the OP.. so let's stop testing each other.. Ok?

I need to get back to my job on-hand shortly.. later my friend.

Best Regards,

Dave :^)

Sorry.. Monday I got called by my boss (I work from home usually) to work on a manual and had to bail. Yesterday, I had to go into work and solve a problem I couldn't solve from home. I left at 9:30 AM Tuesday (to work) and got home at 1:30 AM Wednesday. A 16 hour work day. Just got up two hours ago, for a call from work, then checked in here.

Again.. sorry I can't be on here as much as I like, but my job has my priority. But honestly, I would have solved such by plugging in a real number for A (given A=B) and worked it through and it would have failed somewhere along the procedure (step5?).

Anyway, I admittedly suck at Math. Now.. 45 years ago.. I was top of the class and aced Algebra. But I never use it in the real world. For the type of computer programming I do, Algebra is pretty rare. Even in Electronics it's rare.

Back in College this problem was submitted to the whole School. Given a month to work on it, I was the first to submit an answer Immediately. 60 people submitted answers over the next few weeks, including 10 professors. I was the only person with the correct answer.

Want to try it?

Years later, at a company that makes measuring equipment in electrically noisy environments, my Boss, with a PHD in Mathematics, submitted the ultimate algorithm using Linear Regression for filtering Noise from data gathered and digitized. He included a proof that it was unbeatable. It was a monster algorithm and took huge CPU and Memory Resources to run it.

I beat it with a statistical filter of my own design that took about 30 Bytes of Memory and about 20 lines of code, and run about 1000 times faster. He was furious.

But those were my glory days when I kicked butt against Experts in many fields. Today, I'm a has-been. My last Project (starting soon, before I can retire), requires me to do Voice Analysis and Speech Recognition. The Math on this one will be intense and insane. I have to do it on a small circuit board with a small CPU, little Memory, all self contained in a single stand-alone product. All my current Products Talk now, but this one has to Listen. I'm using a generic DSP. Needless to say.. it is scaring the crap out of me.. but I always succeed when I put my mind to a problem (so far.. lol).

A minor mention.. we have drifted radically from the OP.. so let's stop testing each other.. Ok?

I need to get back to my job on-hand shortly.. later my friend.

Best Regards,

Dave :^)

Posted: **March 1st, 2017, 9:59 pm**

Posted: **March 2nd, 2017, 12:24 am**

I dont quite follow what RT is. But if it's the resistance you're after then it should be 2 (ohms or whatever). The max current will follow the path of least resistance. Lesser amounts will go into the other circuits. Ohm's law from back in college.

Posted: **March 2nd, 2017, 7:07 pm**

Hi Raj,

Less than 2 Ohms, as the "Resistance Total" is a complex function of the whole network.

Example:

1. Two 2-ohm in parallel = 1 ohm.

2. Two 2-ohm in series = 4 ohm.

Anyway, forget about it. Not really relevant to the OP.. as has much of the last number of posts have deviated too.

So.. see you around..

Best wishes,

Dave :^)

Less than 2 Ohms, as the "Resistance Total" is a complex function of the whole network.

Example:

1. Two 2-ohm in parallel = 1 ohm.

2. Two 2-ohm in series = 4 ohm.

Anyway, forget about it. Not really relevant to the OP.. as has much of the last number of posts have deviated too.

So.. see you around..

Best wishes,

Dave :^)

Posted: **March 2nd, 2017, 9:55 pm**

Ok then this is how I have worked it out:

1/RT = 1/2 + 1/(1+2+3) +1/2 +1/2 = 3/2 +1/6 = 10/6 = 5/3

Therefore RT = 3/5 = 0.6 – could be wrong not sure about the second 1/2. In actual practice I would measure it.

Getting back to topic what defines a good theory?

I would say

1. one that assumes very little to explain a lot. example Darwin’s evolution

2, Explains more than it sets out to do. example Newton’s or Einstein’s gravity

1/RT = 1/2 + 1/(1+2+3) +1/2 +1/2 = 3/2 +1/6 = 10/6 = 5/3

Therefore RT = 3/5 = 0.6 – could be wrong not sure about the second 1/2. In actual practice I would measure it.

Getting back to topic what defines a good theory?

I would say

1. one that assumes very little to explain a lot. example Darwin’s evolution

2, Explains more than it sets out to do. example Newton’s or Einstein’s gravity

Posted: **March 2nd, 2017, 10:20 pm**

Hi Raj,

The solution is 1.16666 Ohms.

I brought this problem into a thread about Resistance in water pipe flow.

My solution is found here:

http://www.sciencechatforum.com/viewtopic.php?f=128&t=26667&start=30#p259357

Regards,

Dave :^)

The solution is 1.16666 Ohms.

I brought this problem into a thread about Resistance in water pipe flow.

My solution is found here:

http://www.sciencechatforum.com/viewtopic.php?f=128&t=26667&start=30#p259357

Regards,

Dave :^)

Posted: **March 2nd, 2017, 10:22 pm**

Hi Raj,

Can you start a new thread? I feel bad this one has gone so far off track.

Regards,

Dave :^)

Can you start a new thread? I feel bad this one has gone so far off track.

Regards,

Dave :^)

Posted: **March 2nd, 2017, 10:36 pm**

Seems I was right about the second 1/2 But if that be so, my solution works out to be

1/RT = 7/6

or RT = 6/7 = 0.857 or the inverse of your solution. You sure you are right?

1/RT = 7/6

or RT = 6/7 = 0.857 or the inverse of your solution. You sure you are right?

Posted: **March 2nd, 2017, 11:06 pm**

Hi Raj,

Yes.. of course it's right. I got my picture in the paper because I was the only person to solve it with the correct answer. This pissed off my Math teacher as I had beat him. He then commenced with a campaign to flunk me and I only got a "C" in his class.. all other classes were staright "A"'s. I took this issue to the Dean but he sided with the professor and didn't want to become involved. The Professor promised he would never pass me.. ever. You need a "B" to proceed. So I dropped out, got drafted, and never completed my college education because of this one stupid problem.

Three years later, discounting the null time in the Army, I was Senior R&D Engineer at a major company and would have the <Hire or Not> authority over recent graduates from that school. I climbed the corporate ladder without a degree due to skill and ability that was recognized by my peers.

Thus I got the last laugh in the end. I've been holding the position of "Chief Engineer" at several companies since then.. over the last 25 years. Never had a degree.

Degree's are a silly game in my book.

If it is Old, then it is of little current value.

If it is New, then it is a rush Job and incomplete.

Because technology is changing and growing so fast.

All new-hires have to be trained in specific areas for any given company.

Experience is far more valuable than a Degree.

Again.. for at least the 3rd time.. start a new thread if you want to continue. Or.. perhaps a Mod will be kind enough to Lock this one, as it seems to have been played out anyway and is definitely off track from the OP.

Or.. we get back on track.. and repeat our previous stances. Pick..

Regards,

Dave :^)

Yes.. of course it's right. I got my picture in the paper because I was the only person to solve it with the correct answer. This pissed off my Math teacher as I had beat him. He then commenced with a campaign to flunk me and I only got a "C" in his class.. all other classes were staright "A"'s. I took this issue to the Dean but he sided with the professor and didn't want to become involved. The Professor promised he would never pass me.. ever. You need a "B" to proceed. So I dropped out, got drafted, and never completed my college education because of this one stupid problem.

Three years later, discounting the null time in the Army, I was Senior R&D Engineer at a major company and would have the <Hire or Not> authority over recent graduates from that school. I climbed the corporate ladder without a degree due to skill and ability that was recognized by my peers.

Thus I got the last laugh in the end. I've been holding the position of "Chief Engineer" at several companies since then.. over the last 25 years. Never had a degree.

Degree's are a silly game in my book.

If it is Old, then it is of little current value.

If it is New, then it is a rush Job and incomplete.

Because technology is changing and growing so fast.

All new-hires have to be trained in specific areas for any given company.

Experience is far more valuable than a Degree.

Again.. for at least the 3rd time.. start a new thread if you want to continue. Or.. perhaps a Mod will be kind enough to Lock this one, as it seems to have been played out anyway and is definitely off track from the OP.

Or.. we get back on track.. and repeat our previous stances. Pick..

Regards,

Dave :^)

Posted: **March 3rd, 2017, 12:07 am**

Hi Dave,

I had a look at the link you posted. From a quick perusal I think that analysis is wrong. But I will post on that thread when I have time.

Why is it that only in my posts people ask moderators to shut it down or shunt it off?

Eminent scientific posts like "Backwards Land" or "What the heck is time?" seem to be under no such threats.

Regards

Raj

I had a look at the link you posted. From a quick perusal I think that analysis is wrong. But I will post on that thread when I have time.

Why is it that only in my posts people ask moderators to shut it down or shunt it off?

Eminent scientific posts like "Backwards Land" or "What the heck is time?" seem to be under no such threats.

Regards

Raj

Posted: **March 3rd, 2017, 7:07 am**

As I suspected Dave you have made a mistake. I sat down and worked it out. I used a different method and got 1.130

You have made a mistake in your second diagram.

Not being able to paste my picture but the second 2 ohms should be connected at the end and not the middle. I have posted at your link.

You have made a mistake in your second diagram.

Not being able to paste my picture but the second 2 ohms should be connected at the end and not the middle. I have posted at your link.

Posted: **March 3rd, 2017, 2:11 pm**

Here is the correct configuration: